Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 30

22.8 Electric Potential Due to a Uniformly Charged Arc 753 P P R R R Q + + θ Q + + + + + + + + + + (a) R dθ dq ++ + + + + + ++ s ds (b) + Fig 22.18 (a) A circular arc of radius R, central angle φ, and center P has a uniformly distributed positive charge Q (b) The figure shows how to calculate the electric potential dV at P due to an arc element ds having a charge dq The total electric potential at P due to all elements of the arc is thus: φ V = dθ = kλφ dV = kλ Using Eq 22.52, we get: V = kλφ = k Q R (22.56) This expression is identical to the formula of a point charge The reason for this is that the distance between P and each charge element on the arc does not change and its orientation is irrelevant 22.9 Electric Potential Due to a Uniformly Charged Ring Assume that a ring of radius R has a uniformly distributed total positive charge Q, see Fig 22.19 Additionally, assume that a point P lies at a distance a from the center of the ring along its central perpendicular axis, as shown in the same figure To find the electric potential at P, we first calculate the electric potential dV at P due to a segment of charge dq as follows: dV = k where r = dq r √ R2 + a2 is a constant distance for all elements on the ring (22.57) 754 22 Electric Potential Fig 22.19 A ring of radius R z having a uniformly distributed P positive charge Q The figure Q shows how to calculate the a r electric potential dV at an axial point P due to a segment R of charge dq on the ring dq Thus, the total electric potential at P is: V = k k dq =√ √ R2 + a2 R2 + a2 dV = dq (22.58) Since dq represents the total charge Q over the entire ring, then the total electric potential at P will be given by: V =√ kQ (22.59) R2 + a2 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 The ring P Ring a Charge per unit area shown has a radius r and a Disk radial width dr dr r R To find the electric potential at P, we divide the disk into concentric rings, then calculate the electric potential at P for each ring by using Eq 22.59, and summing up the contribution of all the rings 22.10 Electric Potential Due to a Uniformly Charged Disk 755 Figure 22.20 shows one such ring, with radius r, radial width dr, and surface area dA = 2π rdr Since σ is the charge per unit area, then the charge dq on this ring is: dq = σ dA = 2π rσ dr (22.60) Using this relation in Eq 22.59, and replacing V with dV, R with r, and Q with dq = 2π rσ dr, we can calculate the electric potential resulting from this ring as follows: dV = √ k (2π rσ dr) = π kσ √ r + a2 2rdr r + a2 (22.61) To find the total electric potential, we integrate this expression with respect to the variable r from r = to r = R This gives: R V = (r + a2 )−1/2 (2r dr) dV = π kσ (22.62) To solve this integral, we transform it to the form un du = un+1 /(n + 1) by setting u = r + a2 , and du = 2r dr Thus, Eq 22.62 becomes: u=R2 +a2 R V = π kσ −1/2 (r + a ) (2r dr) = π kσ = π kσ u−1/2 du u=a2 u1/2 1/2 u=R2 +a2 u=a2 = π kσ (R2 + a2 )1/2 1/2 (22.63) − a 1/2 Rearranging the terms, we find that: V = 2π kσ √ R2 + a2 − a (22.64) Using k = 1/4π ◦ , where ◦ is the permittivity of free space, it is sometimes preferable to write this relation as: V = σ √ R + a2 − a ◦ (22.65) 756 22 Electric Potential 22.11 Electric Potential Due to a Uniformly Charged Sphere A solid sphere of radius R has a uniform volume charge density ρ and carries a total positive charge Q First, we find the electric potential in the region r ≥ R by using the electric field obtained in Example 21.9 In this region, we found that E is radial and has a magnitude: Er = k Q r2 (r ≥ R) (22.66) This is the same as the electric field due to a point charge, and hence the electric potential at any point of radius r in this region is given by: Vr = k Q r (r ≥ R) (22.67) The potential at a point on the surface of the sphere (r = R) is: VR = k Q R (22.68) In the region ≤ r ≤ R inside the sphere, we use the result of the electric field obtained → in Example 21.9 In this region, we found that E is radial and has a magnitude: Er = k Q r R3 (0 ≤ r ≤ R) (22.69) For a point that has a radius r in the region ≤ r ≤ R, we can find the potential difference between this point and any point on the surface with a radius R by using → E • d→ s = Er dr in Eq 22.6 Thus: r Vr − VR = − → E • d→ s =− R r Er dr = −k R Q r2 = −k R r=r r=R Q R3 r r dr R (22.70) Q = k (R2 − r ) 2R Using VR = kQ/R in the last result we reach to the following relation: Vr = k r2 Q 3− 2R R (0 ≤ r ≤ R) (22.71) At r = 0, we have V0 = 3kQ/2R, and at r = R, we get VR = kQ/R as expected Figure 22.21 sketches the electric potential in the two regions ≤ r ≤ R and r ≥ R 22.12 Electric Potential Due to a Charged Conductor 757 Fig 22.21 A sketch of the R electric potential V (r) as a r function of r in the two regions ≤ r ≤ R and r ≥ R The curve for the region ≤ r ≤ R is V0 = parabolic and joins smoothly with the curve for the region r2 R2 Q Vr = k r Q V 3k Q 2R Vr = k 2R − 2V r ≥ R, which is hyperbola r R 22.12 Electric Potential Due to a Charged Conductor Assume that a solid conducting sphere of radius R carries a net positive charge Q as shown in Fig 22.22a We found in Chap 21 that the charge on this equilibrium conductor must reside on its outer surface Furthermore, the electric field inside the conductor is zero and the electric field just outside its surface is perpendicular to the surface E Charged conductor Q VR = k R A B + + + R + + + + + V (a) Charged conductor Uniform surface charge density Vr = k R r Q r + ++ + + + + + ++ E + + ++ ++ + ++ Nonuniform surface charge density (b) Fig 22.22 (a) A sketch of the electric potential V (r) as a function of r in the two regions ≤ r ≤ R and r ≥ R for a charged spherical conductor When r = R, the formulas in the regions match (b) For a nonsymmetrical conductor, the surface charge density is greatest at the points where the radius of curvature of the surface is least Consider two arbitrary points A and B on the surface of this spherical conductor, → see Fig 22.22a Since E ⊥ d → s along the path AB on the surface of that conductor, → → then E • d s = Using this result along with Eq 22.6, we find that: 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: dV d a+L d =− 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: d dV =− da da kQa = (R + a2 )3/2 Ea = − √ kQ R2 = −kQ(− 21 )(R2 + a2 )−3/2 (2a) + a2 (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 ◦ σ a = 1− √ 2 ◦ R + a2 Ea = − R + 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=√ R + 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) = 2a kq 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 = αR /(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 Er = Er = kq/r Er = r>c b