HW3-Problem_10 - Version 001 – Homework 3 – Hoffmann...

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Unformatted text preview: Version 001 – Homework 3 – Hoffmann – (56605) Calculate the potential difference between point A and B. Correct answer: 9.5 V. 4 5. VP = 0 2 ke Q 1 ke Q 2 + a+b c 2 ke Q 1 ke Q 1 ke ( Q 1 − Q 2 ) 7. VP = + − a+b b c 2 ke Q 1 8. VP = a 2 ke Q 1 ke Q 2 9. VP = − a+b c ke Q 1 ke Q 2 10. VP = − a+b b Explanation: Using the superposition principle, adding the 3 concentric charge distributions; i.e., Q1 at a, −Q1 at b and Q1 + Q2 at c, gives 6. VP = Explanation: Let : q = 2 C and W = 19 J . The work is W = qV W 19 J V= = = 9. 5 V . q 2C Concentric Conductors 03 p2e1 010 10.0 points Consider a solid conducting sphere with a radius a and charge Q1 on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius b (with b > a) and outer radius c and a net charge Q2 on the shell. Denote the charge on the inner surface of the shell by Q2 and that on the outer surface of the shell by Q2 . Q1 , a b , Q2 P Q2 V= 2 ke Q 1 ke Q 1 ke ( Q 1 + Q 2 ) . − + a+b b c Alternative Solution: V =− =− =− =− Er · dr , by symmetry, Er dr c ∞ c ∞ Er dr − (a+b)/2 Er dr b ke ( Q 1 + Q 2 ) dr r2 − Q2 , c Q1 Assume: The potential at r = ∞ is zero. Find the potential VP at point P , which is the midpoint between a and b. 1. VP = 2. VP = 3. VP = 4. VP = rect 2 ke Q 1 2 ke Q 2 − a+b b 2 ke ( Q 1 − Q 2 ) a+b 2 ke Q 1 a+b 2 ke Q 1 ke Q 1 ke ( Q 1 + Q 2 ) − + cora+b b c = −ke (Q1 + Q2 ) 1 r (a+b)/2 b c ∞ −ke Q1 = 1 r ke Q 1 dr r2 (a+b)/2 b 2 ke Q 1 ke Q 1 ke ( Q 1 + Q 2 ) − + . a+b b c keywords: Charge and Concentric Spheres 011 10.0 points A charge of 2 pC is uniformly distributed ...
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