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Transcript
Electric Potential and Energy The problem: A conducting sphere of radius R and charge +Q, enclosed by a grounded spherical shell (radius 2R) and surrounded by a spherical conductor with inner radius of 3R, outer radius of 4R and the total charge −2Q. • Find the electric potential and the electric field as a function of r • Find the work needed to bring a charge q from ∞ to the center of the system The solution: The boundary conditions φ2R = 0, φ3R = φ4R , Q3 + Q4 = −2Q, are grounded shell conductor given (1) The potentials at the shells are kQ kQ2 kQ3 kQ4 + + + 2R 2R 3R 4R kQ kQ2 kQ3 kQ4 φ3R = + + + 3R 3R 3R 4R kQ kQ2 kQ3 kQ4 φ4R = + + + 4R 4R 4R 4R Solving the equations above, we get (2) φ2R = (3) (4) 1 6 4 Q2R = Q , Q3R = − Q , Q4R = − Q 5 5 5 Substituting the charges into the formulas for the potential of hollow spheres gives kQ , r<R 2R 1 1 kQ( r − 2R ) , R < r < 2R 6 3 φ = kQ( 5r − 5R ) , 2R < r < 3R 1 kQ −5 R , 3R < r < 4R 4kQ − 5r , r > 4R QR = Q , 1 (5) The electric field is the 0 , kQ , r2 r̂ kQ 6 ~ E = 5 r2 r̂ , 0 , 4 kQ − 5 r2 r̂ , derivative of the potential: r<R R < r < 2R 2R < r < 3R 3R < r < 4R r > 4R (6) The work needed to bring a charge +q from ∞ to the center is U = q∆φ = q(φc − φ∞ ) = kQq 2R (7) 2 Electric dipole and capacitors The problem: A cylindrical capacitor has length L. The capacitor consists of two conducting cylinderical shells with a common axis and radii b1 , b2 (b2 > b1 ). 1. What is the capacity of the capacitor? 2. Find the force that acts on the inner cylinder, if it is being pulled along the common axis up to ∆L above the upper part of the capacitor. (∆L L) The solution: 1. The capacity is defined as C = shells using the Gauss’s law Q V. First, we will calculate the field between the cylinderical Q l L 2kQ E = Lr Z b2 2kQ 2kQ b2 ∆φ = − dr = − ln Lr L b1 b1 L Q L = C = = 2π0 b2 b2 V 2k ln b1 ln b1 E2πrl = 4πk (1) (2) (3) (4) 2. Since ∆L is very small, the change of the capacity will also be very small. The energy of a Q2 capacitor is given by: U = 12 CV 2 = 2C . So we presume that the change in the energy will be very small. C − ∆C = 2π0 ∆U F L − ∆L ln bb21 (5) Q2 1 Q2 ∆C 1 1 Q2 ∆L = U (C) − U (C − ∆(C)) = ≈− − = − 2 C C − ∆C 2 C2 2 LC 2 2 ∆U 1Q Q b2 = − = = ln 2 ∆L 2 LC 4π0 L b1 The direction of the force is downwards. 1 (6) (7) Conductors The problem: The inner plate R2 of a cylindrical capacitor is at the potential φ0 , the outer plate R1 is grounded. Between the plates there is charge distribution with ρ = const. Find φ(r) between the plates. Answer: φ= φ0 + kπρ(R22 − R12 ) r ln − kπρ(r2 − R12 ) R1 R ln R2 1 The solution: Let the linear density on the inner plate be λ. Gauss theorem gives Er · 2πrl = 4πk(λ + πρ(r2 − R22 ))l (1) so that Er = 2k(λ − πρR22 ) + 2πρr r (2) The potential will be Z r R1 φ = − Er (r0 )dr0 = 2k(λ − πρR2 ) ln − 2πρ(r2 − R12 ) r R1 Z R2 R1 − 2πρ(R22 − R12 ) φ0 = − Er (r0 )dr0 = 2k(λ − πρR2 ) ln R 2 R1 Find λ from the second equation and substitute into the first. 1 (3) (4)