22.4 Electric Potential Due to a Point Charge 743 bringing q1 by itself from infinity and putting it in place as shown in Fig 22.9a, we no work for such a move The electric potential at point P which is at a distance r12 from q1 is given by Eq 22.26 as VP = kq1 /r12 Later, by bringing q2 without acceleration from infinity to point P at a distance r12 from q1 , as shown in Fig 22.9b, we must work W∞P (app) for such a move, since q1 exerts an electrostatic force on q2 during the move P VP k r12 q1 r12 q2 r12 U k q1 q1 (a) q q2 r12 (b) Fig 22.9 (a) The potential VP at a distance r12 from a point charge q1 (b) The potential energy of two point charges is U = kq1 q2 /r12 We can calculate W∞P (app) by using K = W∞P (app) + W∞P = 0, i.e W∞P (app) = −W∞P When we replace q2 by the general charge q in Eq 22.6, we find that: VP − V∞ = UP − U∞ W∞P =− q2 q2 (22.28) Setting V∞ as well as U∞ to zero (our reference point at ∞), we get: VP = UP W∞P (app) = q2 q2 ⇒ UP = W∞P (app) = q2 VP (22.29) Substituting with VP = kq1 /r12 , we can generalize the electric potential energy of a system of two point charges q1 and q2 separated by a distance r12 as follows: U=k q1 q2 r12 (22.30) If the charges have the same sign, we have to positive work to overcome their mutual repulsion, and then U is positive If the charges have opposite signs, we have 744 22 Electric Potential to negative work against their mutual attraction to keep them stationary, and then U is negative When the system consists of more than two charges, we calculate the potential energy of each pair and add them algebraically For instance, the total potential energy of three charges q1 , q2 , and q3 is: q1 q2 q1 q3 q2 q3 + + r12 r13 r23 U=k (22.31) Example 22.5 Two charges q1 = µC and q2 = − µC are fixed in their positions and separated by a distance d = 10 cm; see Fig 22.10a (a) Find the total electric potential at the point P in Fig 22.10a (b) Find the change in potential energy of the two charges when a third charge q3 = µC is brought from ∞ to P, see Fig 22.10b (c) What is the total electric potential energy of the three charges? P + d q1 + d d d d - q2 q1 + (a) q3 d - q2 (b) Fig 22.10 Solution: (a) For two charges, Eq 22.27 gives: VP = k × 10−6 C −4 × 10−6 C q2 q1 + = (9 × 109 N.m2 /C2 ) + d d 0.1 m 0.1 m = −1.8 × 105 V (b) When we replace q3 by the charge q of Eq 22.6, and bring q3 from infinity to point P, this equation gives: U = q3 (VP − V∞ ) = (6 × 10−6 C)(−1.8 × 105 V − 0) = −1.08 J 22.4 Electric Potential Due to a Point Charge 745 (c) The total electric potential energy of the three charges is: q1 q3 q2 q3 q1 q2 + + d d d [(2)(−4) + (2)(6) + (−4)(6)] × 10−12 C2 = (9 × 109 N.m2 /C2 ) = −1.8 J 0.1 m U=k 22.5 Electric Potential Due to a Dipole As introduced in Chap 20, an electric dipole consists of a positive charge q+ = +q and an equal-but-opposite negative charge q− = −q separated by a distance 2a Let us find the electric potential at a point P in the xy-plane as shown in Fig 22.11a We assume that V+ is the electric potential produced at P by the positive charge, and that V− is the electric potential produced at P by the negative charge y y P(x,y) r r r r q - ( a,0) r r r y q + O (a,0) xa x a q x q - + O x 2a (b) (a) Fig 22.11 (a) The electric potential V at point P(x, y) due to an electric dipole located along the x-axis with a length 2a Point P is at a distance r from the midpoint O of the dipole, where OP makes an angle θ with the dipole’s x-axis (b) When P is very far, the lines of length r+ and r− are approximately parallel to the line OP The total electric potential at P is thus: V = V+ + V− = k q+ q− 1 +k = kq − r+ r− r+ r− = kq r− − r+ r+ r− (22.32) = (x − a)2 + y2 and r = (x + a)2 + From the geometry of Fig 4.11a, we find that r+ − y2 Accordingly, Eq 22.32 becomes: V = kq (x − a)2 + y2 − (x + a)2 + y2 (22.33) 746 22 Electric Potential Note that V = ∞ at P(x = a, y = 0) and V = −∞ at P(x = −a, y = 0) Figure 22.12 shows a plot of the general shape of V in the xy-plane Fig 22.12 A computer- V y generated plot of the electric potential V in the xy-plane for an electric dipole The predicted infinite values of V are not plotted x Because naturally occurring dipoles have very small lengths, such as those possessed by many molecules, we are usually interested only in points far away from the dipole, i.e r 2a Considering these conditions, we find from Fig 22.11b that: r− − r+ 2a cos θ, and r− r+ r2 (22.34) When substituting these approximate quantities in Eq 22.32, we find that: V = kq 2a cos θ r2 (r 2a) (22.35) As introduced in Chap 20, the product of the positive charge q and the length of the dipole 2a is called the magnitude of the electric dipole moment, p = 2aq The direction of → p is taken to be from the negative charge to the positive charge of the → → dipole, i.e p = p i This indicates that the angle θ is measured from the direction of → p Using this definition, we have: V =k p cos θ r2 (r 2a) (22.36) Example 22.6 Find the electric potential along the axis of the electric dipole at the four points A, B, C, and D in Fig 22.13 22.5 Electric Potential Due to a Dipole 747 y D -q C - B +q A + x O (-a,0) (a,0) Fig 22.13 Solution: We use Eq 22.32 with r+ and r− as the distance from each point to the positive and negative charges, respectively: (1) For point A in Fig 22.13, we have x > a Therefore, r+ = x − a and r− = a + x The electric potential VA is: 1 1 − = kq − r+ r− x−a a+x 2k qa (x a) x2 VA = kq = 2k qa x − a2 (VA positive) (2) For point B in Fig 22.13, we have < x < a Therefore r+ = a − x and r− = a + x The electric potential VB is: VB = kq 1 − r+ r− = kq 1 − a−x a+x = 2kq x a2 − x (VB positive) (3) For point C in Fig 22.13, we have −a < x < Therefore r+ = a − x and r− = a + x The electric potential VC is: VC = kq 1 − r+ r− = kq 1 − a−x a+x = 2kq x a2 − x (VC negative) (4) For point D in Fig 22.13, we have x < −a Therefore r+ = a − x and r− = −x − a The electric potential VD is: 1 − r+ r− 2kqa − (x x VD = kq 22.6 = kq 1 + a−x a+x =− 2kqa − a2 x2 −a) (VD negative) Electric Dipole in an External Electric Field Consider an electric dipole of electric dipole moment → p is placed in a uniform → external electric field E , as shown in Fig 22.14 Do not get confused between the 748 22 Electric Potential field produced by the dipole and this external field In addition, we assume that the → vector → p makes an angle θ with the external field E +q + 2a p o −F θ F E p ⊗ θ E - -q (b) (a) Fig 22.14 (a) An electric dipole has an electric dipole moment → p in an external uniform electric → → field E The angle between → p and E is θ The line connecting the two charges represents their rigid connection and their center of mass is assumed to be midway between them (b) Representing the electric → dipole by a vector → p in the external electric field E and showing the direction of the torque → τ into the page by the symbol ⊗ → → Figure 22.14 shows a force F , of magnitude qE in the direction of E , is exerted on → the positive charge, and a force −F , of the same magnitude but in opposite direction, is exerted on the negative charge The resultant force on the dipole is zero, but since the two forces not have the same line of action, they establish a clockwise torque τ→ about the center of mass of the two charges at o The magnitude of this torque about o is: τ = (2a sin θ )F (22.37) Using F = qE and p = 2aq, we can write this torque as: τ = (2aq sin θ )E = pE sin θ (22.38) The vector torque τ→ on the dipole is therefore the cross product of the vectors → p → and E Thus: → τ→ = → p ×E (22.39) The effect of this torque is to rotate the dipole until the dipole moment → p is aligned → with the electric field E 22.6 Electric Dipole in an External Electric Field 749 Potential energy can be associated with the orientation of an electric dipole in an electric field The dipole has the least potential energy when it is in the equilibrium → orientation, which occurs when → p is along E On the other hand, the dipole has the → greatest potential energy when → p is antiparallel to E We chose the zero-potential→ energy configuration when the angle between → p and E is 90◦ According to Eq 22.3, U = UB − UA = −WAB , we can find the electric potential energy of the dipole by calculating the work done by the field from the initial orientation θ = 90◦ , where UA ≡ U(90◦ ) = 0, to any orientation θ, where UB ≡ U(θ ) In addition, we use the relation W = τ dθ, to find U(θ ) as follows: U(θ ) − U(90◦ ) = −W90◦ →θ = − θ τ dθ (22.40) 90◦ Letting U(θ ) ≡ U, U(90◦ ) = 0, τ = pE sin θ, and integrating we get: U = −pE cos θ (22.41) This relation can be written in vector form as follows: → U = −→ p •E (22.42) Equation 22.42 shows the least and greatest value of U as follows: 180° p τ max p 90° τ E p 22.7 E E Electric Potential Due to a Charged Rod For a Point on the Extension of the Rod Figure 22.15 shows a rod of length L with a uniform positive charge density λ and a total charge Q In this figure, the rod lies along the x-axis and point P is taken to be at the origin of this axis, located at a distance a from the left end When we consider a segment dx on the rod, the charge on this segment will be dq = λ dx 750 22 Electric Potential y P dx x + + + + + + + + + + + + dq L a x Fig 22.15 The electric potential V at point P due to a uniformly charged rod lying along the x-axis The electric potential due to a segment of charge dq at a distance x from P is k dq/x The total electric potential is the algebraic sum of all the segments of the rod The electric potential dV at P due to this segment is given by: dV = k dq λ dx =k x x (22.43) We obtain the total electric potential at P due to all the segments of the rod by integrating from one end of the rod (x = a) to the other (x = a + L) as follows: a+L V = dV = k a λ dx = kλ x a+L x −1 dx = kλ| ln x|a+L = kλ {ln(a + L) − ln a} a a Therefore: V = kλ ln a+L a = kQ a+L ln L a (22.44) For a Point on the Perpendicular Bisector of the Rod A rod of length L has a uniform positive charge density λ and a total charge Q The rod is placed along the x-axis as shown in Fig 22.16 Assuming that P is a point on the perpendicular bisector of the rod and is located a distance a from the origin of the x-axis, then the charge of a segment dx on the rod will be dq = λ dx The electric potential dV at P due to this segment is: dV = k λ dx dq =k r r (22.45) The total electric potential at P due to all segments of the rod is given by two times the integral of dV from the middle of the rod (x = 0) to one of its ends (x = L/2) Thus: 22.7 Electric Potential Due to a Charged Rod 751 Fig 22.16 A rod of length L y with a uniform positive charge P density λ and an electric potential dV at point P due to a a charge segment r dq + + + + + + + + + ++ + x dx L x=L/2 V =2 L/2 dV = 2kλ x=0 dx r x (22.46) To perform the integration of this expression, we relate the variables x and r From √ the geometry of Fig 22.16, we use the fact that r = x + a2 Therefore, Eq 22.46 becomes: L/2 V = 2kλ dx (x + a2 )1/2 (22.47) From the table of integrals in Appendix B, we find that: (x dx = ln(x + + a2 )1/2 V = 2kλ ln(x + x + a2 ) Thus: = 2kλ ln(L/2 + Therefore: V = 2kλ ln x + a2 ) (22.48) L/2 (L/2)2 + a2 ) − ln(a) L/2 + (L/2)2 + a2 a (22.49) When we use the fact that the total charge Q = λL, we get: V = L/2 + 2kQ ln L (L/2)2 + a2 a (22.50) 752 22 Electric Potential For a Point Above One End of the Rod When a point P is located at a distance a from one of the rod’s ends, see Fig 22.17, we can perform similar calculations to find that: L+ kQ ln V = L √ L + a2 a Fig 22.17 A setup similar to y Fig 22.16 except P is above P one end a 22.8 (22.51) r dq ++++ +++++ +++ x dx L x Electric Potential Due to a Uniformly Charged Arc Assume that a rod has a uniformly distributed total positive charge Q Assume now that the rod is bent into an arc of radius R and central angle φ rad, see Fig 22.18a To find the electric potential at the center P of this arc, we first let λ represent the linear charge density of this arc, which has a length Rφ Thus: λ= Q Rφ (22.52) For an arc element ds subtending an angle dθ at P, we have: ds = R dθ (22.53) Therefore, the charge dq on this arc element will be given by: dq = λ ds = λ R dθ (22.54) To find the electric field at P, we first calculate the differential electric potential dV at P due to the element ds of charge dq, see Fig 22.18b, as follows: dV = k dq = kλ dθ R (22.55) 758 22 Electric Potential B VB − VA = − → E • d→ s =0 (22.72) A During equilibrium, the electric potential V is constant everywhere on the surface of this charged spherical conductor and equal to VR = kQ/R, or VR = 4π kRσ in terms of the surface charge density σ Furthermore, because the electric field is zero inside the conductor, the electric potential would be constant everywhere inside the conductor and is equal to its value at the surface Outside this spherical conductor, the electric potential is Vr = kQ/r for r ≥ R Figure 22.22a plots V against r and shows the dependence of V (r) on r for the whole range of r If the conductor is not symmetric as in Fig 22.22b, the electric potential is constant everywhere on its surface, but the surface charge density is not uniform Since V = const and V ∝ Rσ, i.e Rσ = const., the surface charge density increases as the radius of curvature decreases 22.13 Potential Gradient We defined the potential difference between two points A and B as the negative of → the work done by the electric field E per unit charge in moving the charge from A to B, see Eq 22.6 Thus: B VB − VA = − → E • d→ s (22.73) A If we write VB − VA = B A dV = − B→ A E • d→ s , then we must have: → dV = −E • d → s (22.74) → If the electric field has only one component Ex along the x-axis, then E • d → s = Ex dx The last equation becomes dV = −Ex d x, or: Ex = − dV dx (22.75) 22.13 Potential Gradient 759 Thus, the x component of the electric field is equal to the negative of the derivative of the electric potential with respect to x → If the field is radial, i.e V = V (r) as introduced in Sect 22.4, then E • d → s = Er dr and we can express Eq 22.74 as: Er = − → → → dV dr → (22.76) → → → Generally, E = Ex i + Ey j + Ez k and d → s = dx i + dy j + dz k Then: → dV = −E • d → s = −Ex dx − Ey dy − Ez dz (22.77) When V = V (x, y, z), the chain rule of differentiation gives: dV = ∂V ∂V ∂V dx + dy + dz ∂x ∂y ∂z (22.78) By comparing the last two equations, we get the potential gradients: Ex = − ∂V ∂x Ey = − ∂V ∂y Ez = − ∂V ∂z (22.79) Example 22.7 From the formulas of the electric potential given by Eqs 22.26, 22.44, 22.59, and 22.65, find the formulas of the electric fields Solution: To get the electric field from the electric potential, we use Eqs 22.76 or 22.79 depending on the system coordinates (1) From the point-charge formula given by Eq 22.26, we have a radial electric field Thus: Er = − d q dr −1 q dV =− k = −kq =k dr dr r dr r (Identical to Eq 20.4) (2) From the charged-rod formula given by Eq 22.44, our variable is the distance a from the end of the rod Thus: d a+L d dV =− kλ ln = −kλ {ln(a + L) − ln a} da da a da kλL − = (Identical to Eq 20.26) = −kλ a+L a a(a + L) Ea = − 760 22 Electric Potential (3) From the charged-ring formula given by Eq 22.59, our variable is the distance a from the center of the ring Thus: dV d =− da da kQa = (R + a2 )3/2 Ea = − kQ √ R2 + a2 = −kQ(− 21 )(R2 + a2 )−3/2 (2a) (Identical to Eq 20.50) (4) From the charged-disk formula given by Eq 22.65, our variable is the distance a from the center of the disk Thus: d σ dV =− da da ◦ a σ 1− √ = 2 ◦ R + a2 Ea = − R2 + a2 − a =− σ ◦ (R2 + a2 )−1/2 (2a) − (Identical to Eq 20.56) The expressions that we have arrived at for the electric potentials established by simple charge distributions are presented in Table 22.1 Table 22.1 Electric potential due to simple charge distributions Charge distribution Two oppositely charged conducting plates separated by a distance d Electric potential V = VB − VA = − Ed Along the field V = VB − VA = Ed Opposite the field q r Single point charge q V =k Charged ring of radius R with a uniformly distributed total charge Q kQ V=√ R2 + a2 Disk of radius R having a uniform surface charge density σ V= Charge q uniformly distributed on the surface of a conducting sphere of radius R V = r >0 a≥0 σ √ R + a2 − a ◦ ⎧ q ⎪ k ⎪ ⎨ r r ≥R ⎪ ⎪ ⎩k q R r ≤R a>0 22.14 The Electrostatic Precipitator 761 22.14 The Electrostatic Precipitator Electrostatic precipitators are highly efficient filtration devices used to remove particles from a flowing gas (such as air) They this using the force of an induced electrostatic charge Such devices remove particulate matter from combustion gases, and as a result reduce air pollution Most precipitators on the market today are capable of eliminating more than 99% of the ash from smoke A schematic diagram of an electrostatic precipitator is shown in Fig 22.23 Applied between the central wire and the duct walls, where smoke is flowing up the duct, is a large voltage of several thousand volts (50–100 kV) To generate an electric field that is directed toward the wire, the wire is maintained at a negative electric potential with respect to the walls Such a large electric field produces a discharge around the wire, which causes the air near the wire to contain electrons, positive ions, and negative ions such as O− Insulator Without Precipitator Cleaned air Polluted air weight Dirt Dirt out Fig 22.23 (a) A schematic diagram of an electrostatic precipitator The high negative electric potential on the central wire creates a discharge in the vicinity of the wire, causing dirt to fall down (b) Pollution from a power-plant’s chimney not equipped with an electrostatic precipitator Polluted air enters the duct from the bottom and moves near the coiled wire The discharge creates electrons and negative ions, which accelerate toward the outer wall due to the force of the electric field Consequently, the dirt particles become charged by collisions and ion capture Because most of the charged dirt particles are negative, 762 22 Electric Potential they are drawn to the walls by the electric field Periodic shaking of the duct loosens the particles, which are then collected at the bottom 22.15 The Van de Graaff Generator When a charged conductor is connected to the inside of a hollow conductor, all the charge is transferred to the outer surface of the hollow conductor regardless of any charge already retained by the conductor The generator invented by Robert Van de Graaff makes use of this principle, where a “conveyor belt” carries out the charge continuously, see the schematic diagram of Fig 22.24a A F B E D C http://www.explainthatstuff.com/electricity.html (a) (b) Fig 22.24 (a) The charge in the Van de Graaff generator is deposited at E and transferred to the dome at F (b) By touching the dome, each hair strand becomes charged and repels strands around it This generator consists of a hollow metallic dome A supported by an insulating stand B mounted on a grounded metal base C and a non-conducting belt D running over two non-conducting pulleys The belt is charged as a result of the discharge produced by the metallic needle at E, which is maintained at a positive electric potential of about 104 V The positive charge on the moving belt is transferred to the dome by the needle at F, regardless of the dome’s electric potential It is possible to increase the potential of the dome until electrical ionization occurs in the air Since the ionization breakdown of air occurs at an electric field of about × 106 V/m, a sphere of m can be raised to maximum of Vmax = ER = (3 × 106 V/m)(1 m) = × 106 V The dome’s electric potential can be increased further by placing the dome in vacuum and by increasing the radius of the sphere 22.16 Exercises 763 22.16 Exercises Section 22.1 Electric Potential Energy (1) A charge q = 2.5 × 10−8 C is placed in an upwardly uniform electric field of magnitude E = × 104 N/C What is the change in the electric potential energy of the charge-field system when the charge is moved (a) 50 cm to the right? (b) 80 cm downwards? (c) 250 cm upwards at an angle 30◦ from the horizontal? (2) Redo question to calculate the work done on the charge q by the electric field (3) Redo question to calculate the work done on the charge q by an external agent such that the charge moves in each case without changing its kinetic energy Section 22.2 Electric Potential (4) How much work is done by an external agent in moving a charge q = −9.63 × 104 C from a point A where the electric potential is 10 V to a point B where the electric potential is −4 V? How many electrons are there in this charge? Is this number related to any of the known physical constants? Section 22.3 Electric Potential in a uniform Electric Field (5) The electric potential difference between the accelerating plates in the electron gun of a TV tube is 5,550 V, while the separation between the plates is d = 1.5 cm Find the magnitude of the uniform electric field between the plates (6) An electron moves a distance d = cm when released from rest in a uniform electric field of magnitude E = × 104 N/C (a) What is the electric potential difference through which the electron has passed? (b) Find the electron’s speed after it has moved that distance? (7) Two large parallel metal plates are oppositely charged with a surface charge density of magnitude σ = 1.2 nC/m2 , see Fig 22.25 (a) Find the electric field between the plates (b) If the electric potential difference between these two plates is 10 V, what is the distance between the plates? (8) A uniform electric field of magnitude × 102 N/C is directed in the positive x direction as shown in Fig 22.26 In this figure, the coordinates of point A are (0.3, −0.2) m and the coordinates of point B are (−0.5, 0.4) m Calculate the potential difference VB − VA using: (a) the path A → C → B (b) the direct path A → B 764 22 Electric Potential Fig 22.25 See Exercise (7) + +σ −σ + E + + Fig 22.26 See Exercise (8) y E B C x A (9) An insulated rod has a charge Q = 20 µC and a mass m = 0.05 kg The rod is released from rest at a location A in a uniform electric field of magnitude 104 N/C directed perpendicular to the field, see Fig 22.27 and neglect gravity (a) Find the speed of the rod when it reaches location B after it has traveled a distance d = 0.5 m (b) Does the answer to part (a) change when the rod is released at an angle θ = 45◦ relative to the electric field? Fig 22.27 See Exercise (9) B A A= m E + + + + + + Q + + + + + + Q B m d Section 22.4 Electric Potential Due to a Point Charge (10) (a) What is the electric potential at a distances and cm, from a proton? (b) What is the potential difference between these two points? 22.16 Exercises 765 (11) Redo Exercise 10 for an electron (12) At a distance r from a particular point charge q, the electric field is 40 N/C and the electric potential is 36 V Determine: (a) the distance r, (b) the magnitude of the point charge q (13) Two point charges q1 = +2 µC and q2 = −6 µC are separated by a distance L = 12 cm, see Fig 22.28 Find the point at which the resultant electric potential is zero Fig 22.28 See Exercise (13) q1 q2 L + - (14) Two charges q1 = −2 µC and q2 = +2 µC are fixed in their positions and separated by a distance d = 10 cm, see the top part of Fig 22.29 (a) What is the electric field at the origin due these two charges? (b) What is the electric potential at the origin and the electric potential energy of the two charges? (c) Find the change in potential energy of the two charges when a third charge q3 = µC is brought from ∞ to O, see the bottom part of Fig 22.29 Fig 22.29 See Exercise (14) q2 q1 - + x O d q1 q3 q2 - + O d + x (15) Three equal charges q1 = q2 = q3 = nC are located at the vertices of an equilateral triangle of side a = cm, see Fig 22.30 Find the electric potential at point P, which is at the center of the base of the triangle Fig 22.30 See Exercise (15) + q3 a q1 + a P a + q2 766 22 Electric Potential (16) Three negative point charges are placed at the vertices of an isosceles tri√ angle as shown in Fig 22.31 Given that a = cm, q1 = q3 = −2 nC, and q2 = −4 nC, find the electric potential at point P (which is midway between q1 and q3 ) Fig 22.31 See Exercise (16) q1 P a q2 - - q3 a Section 22.5 Electric Potential Due to a Dipole (17) An electric dipole is located along the x-axis with its center at the origin The dipole has a negative charge −q at (−a, 0) and a positive charge +q at (+a, 0), see Fig 22.11 Show that the electric potential on the y-axis is zero for any value of y (18) Assume that a third positive charge +q is placed at the origin of the dipole of Fig 22.11, so that the new configuration will be as show in Fig 22.32 Show that the electric potential for far away points (such as P) on the dipole axis is given by: V (x) = kq 2a 1+ x x −q +q +q - + + -a +a a) P x Fig 22.32 See Exercise (18) (x x 22.16 Exercises 767 (19) The permanent electric dipole moment of the ammonia molecule NH3 is p = 4.9 × 10−30 C.m Find the electric potential due to an ammonia molecule at a distance 52 nm away along the axis of the dipole Section 22.6 Electric Potential Due to a Charged Rod (20) A non-conductive rod has a uniform positive charge density +λ, a total charge Q along its right half, a uniform negative charge density −λ, and a total charge −Q along its left half, see Fig 22.33 (a) What is the electric potential at point A? (b) What is the electric potential at point B? Fig 22.33 See Exercise (20) y B -Q +Q y A x + + + + + + L /2 2L Section 22.7 Electric Potential Due to a Uniformly Charged Arc (21) A non-conductive rod has a uniformly distributed charge per unit length −λ The rod is bent into a circular arc of radius R and central angle 120◦ , see Fig 22.34 Find the electric potential at the center of the arc Fig 22.34 See Exercise (21) P R R 120° −λ (22) A non-conductive rod has a uniformly distributed charge per unit length λ Part of the rod is bent into a semicircular arc of radius R and the rest is left as two straight rod segments each of length R as shown in Fig 22.35 Find the electric potential at point P 768 22 Electric Potential λ Fig 22.35 See Exercise (22) λ λ R R P R Section 22.8 Electric Potential Due to a Uniformly Charged Ring (23) A uniformly charged insulated rod of charge Q = −8 µC and length L = 15.0 cm is bent into the shape of a circle Find the electric potential at the center of the circle If the rod is bent into the shape of a semicircle, find the electric field at its center (24) A ring of radius R has a uniformly distributed total positive charge Q, see Fig 22.19 Find the point on the axis of the ring where the electric potential is half the value of the electric potential at the center (25) An annulus of inner radius R1 and outer radius R2 has a uniform surface charge per unit area σ Calculate the electric potential at the point P which lies at a distance a from the center of the annulus along its central axis, see Fig 22.36 Fig 22.36 See Exercise (25) z σ P a R2 R1 Section 22.9 Electric Potential Due to a Uniformly Charged Disk (26) The disk of Fig 22.20 has a radius R = cm If its surface charge density is µC/m2 from r = to R/2 and 1.5 µC/m2 from r = R/2 to R Find the electric potential at point P on the central axis of the disk, at a distance a = R/2 from its center 22.16 Exercises 769 (27) Show that if the disk of Fig 22.20 has a radius R and a fixed charge Q, the potential on the z-axis reduces to that of a point charge at the origin in the limit R/a → 0, i.e far away from the disk along the z-axis (28) Assume that a disk of radius R has a non-uniform surface charge density σ = αr, where α is a constant and r is the distance from the center of the disk, see Fig 22.37 Find the electric potential at point P on the central axis of the disk, at a distance a from its center (Hint: use the electric potential produced by an element in the form of a ring of radius r and thickness dr.) Fig 22.37 See Exercise (28) z Charge per unit area σ = αr dr P Ring a r Disk R Section 22.10 Potential Due to a Uniformly Charged Sphere (29) A charge Q is distributed uniformly throughout a spherical volume of radius R, see Fig 22.21 (a) Find the point where the electric potential is half the value of the electric potential at the center (b) What is the potential difference between a point on the surface and the sphere’s center? (30) The charge density inside a non-conductive sphere of radius R varies as ρ = αr (C/m3 ), where α is a constant and r is the radial distance from the center of the sphere The electric fields inside and outside the sphere are radial by symmetry and given by: E = αr /4 for r ≤ R ◦ E = αR4 /(4 ◦r ) for r ≥ R (a) Find the electric potential inside the sphere (b) Find the electric potential outside the sphere 770 22 Electric Potential Section 22.11 Electric Potential Due to a Charged Conductor (31) A spherical conductor has a radius R = 10 cm and a positive charge of 20 µC Find the electric potential at: (a) cm, (b) cm, (c) 10 cm, and (d) 15 cm from its center (32) An initially uncharged spherical conductor has a radius R = 20 cm (a) How many electrons should be removed from the sphere to produce an electrical potential of × 106 V at its surface? (b) At that state, what is its surface charge density? (33) A metal sphere of radius a has a charge q and is placed at the center of a hollow metal sphere of inner radius b and outer radius c, which carries a charge Q, see Fig 22.38 (a) Use Gauss’s law to show that: Er = k(q + Q)/r r>c Er = b[...]... potential due to an ammonia molecule at a distance 52 nm away along the axis of the dipole Section 22 .6 Electric Potential Due to a Charged Rod (20 ) A non-conductive rod has a uniform positive charge density +λ, a total charge Q along its right half, a uniform negative charge density −λ, and a total charge −Q along its left half, see Fig 22 .33 (a) What is the electric potential at point A? (b) What... used to remove particles from a flowing gas (such as air) They do this using the force of an induced electrostatic charge Such devices remove particulate matter from combustion gases, and as a result reduce air pollution Most precipitators on the market today are capable of eliminating more than 99% of the ash from smoke A schematic diagram of an electrostatic precipitator is shown in Fig 22 .23 Applied... dV d =− da da kQa = 2 (R + a2 )3 /2 Ea = − kQ √ R2 + a2 = −kQ(− 21 )(R2 + a2 )−3 /2 ( 2a) (Identical to Eq 20 .50) (4) From the charged-disk formula given by Eq 22 .65, our variable is the distance a from the center of the disk Thus: d σ dV =− da da 2 ◦ a σ 1− √ = 2 2 ◦ R + a2 Ea = − R2 + a2 − a =− σ 2 ◦ 1 2 (R2 + a2 )−1 /2 ( 2a) − 1 (Identical to Eq 20 .56) The expressions that we have arrived at for the electric... field Periodic shaking of the duct loosens the particles, which are then collected at the bottom 22 .15 The Van de Graaff Generator When a charged conductor is connected to the inside of a hollow conductor, all the charge is transferred to the outer surface of the hollow conductor regardless of any charge already retained by the conductor The generator invented by Robert Van de Graaff makes use of this... where a “conveyor belt” carries out the charge continuously, see the schematic diagram of Fig 22 .2 4a A F B E D C http://www.explainthatstuff.com/electricity.html (a) (b) Fig 22 .24 (a) The charge in the Van de Graaff generator is deposited at E and transferred to the dome at F (b) By touching the dome, each hair strand becomes charged and repels strands around it This generator consists of a hollow metallic... ring, then the total electric potential at P will be given by: kQ V =√ R2 + a2 (22 .59) 22 .10 Electric Potential Due to a Uniformly Charged Disk Assume that a disk of radius R has a uniform positive surface charge density σ, and a point P lies at a distance a from the disk along its central perpendicular axis, see Fig 22 .20 Fig 22 .20 A disk of radius R z has a uniform positive surface charge density σ... spherical conductor has a radius R = 20 cm (a) How many electrons should be removed from the sphere to produce an electrical potential of 3 × 106 V at its surface? (b) At that state, what is its surface charge density? (33) A metal sphere of radius a has a charge q and is placed at the center of a hollow metal sphere of inner radius b and outer radius c, which carries a charge Q, see Fig 22 .38 (a) Use Gauss’s... potential is half the value of the electric potential at the center (25 ) An annulus of inner radius R1 and outer radius R2 has a uniform surface charge per unit area σ Calculate the electric potential at the point P which lies at a distance a from the center of the annulus along its central axis, see Fig 22 .36 Fig 22 .36 See Exercise (25 ) z σ P a R2 R1 Section 22 .9 Electric Potential Due to a Uniformly... Charged Disk (26 ) The disk of Fig 22 .20 has a radius R = 4 cm If its surface charge density is 2 µC/m2 from r = 0 to R /2 and 1.5 µC/m2 from r = R /2 to R Find the electric potential at point P on the central axis of the disk, at a distance a = R /2 from its center 22 .16 Exercises 769 (27 ) Show that if the disk of Fig 22 .20 has a radius R and a fixed charge Q, the potential on the z-axis reduces to that... integrate this expression with respect to the variable r from r = 0 to r = R This gives: R V = (r 2 + a2 )−1 /2 (2r dr) dV = π kσ (22 . 62) 0 To solve this integral, we transform it to the form un du = un+1 /(n + 1) by setting u = r 2 + a2 , and du = 2r dr Thus, Eq 22 . 62 becomes: u=R2 +a2 R V = π kσ 2 −1 /2 (r + a ) 2 (2r dr) = π kσ 0 = π kσ u−1 /2 du u =a2 u=R2 +a2 u1 /2 1 /2 u =a2 = π kσ (R2 + a2 )1 /2 1 /2 (22 .63)