6b-e-potential

# 6b-e-potential - Electrostatic Potential: There are three...

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Unformatted text preview: Electrostatic Potential: There are three problems in the “spherical symmetry” section and three problems in the “cylindrical symmetry” section. Each problem in the “spherical symmetry” section is exactly analogous to the corresponding problem in the “cylindrical symmetry” section. Suppose we have a single charge and two points in space, B and A, where point B is closer to the charge than point A. If the charge is positive , then point B is at higher potential than point A. If the charge is negative , then point A is at higher potential than point A. In general, potential increases if we move opposite the direction of the electric field lines, and decreases if we move in the same direction as the field lines. Spherical Symmetry Problem 1 (straightforward): A thin spherical insulating shell has radius a and positive charge Q uniformly distributed over its surface. 1: What amount of work is required to move a positive point charge q from point A to point B? Point A is outside the shell, at a radial distance r i > a from the shell’s center, while point B is inside the shell, at a radial distance r f < a from the shell’s center. 2: With zero potential defined at r = ∞ , what is the electrostatic potential at r = r i ? 3: With zero potential defined at r = ∞ , what is the electrostatic potential at r = r f ? & ¡ ¢ £ ¡ ¤ ¥ 1 Solution: Let r denote radial distance from the center of the shell. The fields inside and outside the shell are E ( r ; r < a ) = 0 E ( r ; r ≥ a ) = Q 4 π² r 2 (1) We now compute the change in potential in going from point A, outside the shell, to point B, inside the shell. This is given by integrating over the electric field from the initial to the final position. Δ V AB =- Z r f r i dr E ( r ) =- Z a r i dr E ( r ; r ≥ a )- Z r f a dr E ( r ; r < a ) =- Z a r i dr Q 4 π² r 2- Z r f a dr (0) = Q 4 π² 1 a- 1 r i ¶- = Q 4 π² 1 a- 1 r i ¶ (3) The amount of work required to move a positive point charge q from A to B is therefore W AB = qV AB = Qq 4 π² 1 a- 1 r i ¶ (4) We now turn to part 2 . With zero potential at r = ∞ , the electrostatic potential at r = r i > a is V ( r i ) =- Z r i ∞ dr E ( r ) =- Z r i ∞ dr E ( r ; r ≥ a ) =- Z r i ∞ dr Q 4 π² r 2 = Q 4 π² r i (5) We now turn to part 3 . With zero potential at r = ∞ , the electrostatic potential at r = r f < a is 2 V ( r f ) =- Z r f ∞ dr E ( r ) =- Z a ∞ dr E ( r ; r ≥ a )- Z r f a dr E ( r ; r < a ) =- Z a ∞ dr Q 4 π² r 2- Z r f a dr (0) =- Z a ∞ dr Q 4 π² r 2 = Q 4 π² a (6) Note that the change in potential given by equation (3) is just Δ V AB = V ( r f )- V ( r i ) (7) Problem 2 (medium): Consider two concentric spherical insulating shells, where the inner shell has radius a and the outer shell has radius b , where b > a . The inner shell has a positive charge Q a uniformly distributed over its surface, while the outer shell has a positive charge Q b uniformly distributed over its surface. A positive point chargeuniformly distributed over its surface....
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## This note was uploaded on 01/29/2010 for the course PHYSICS 6B 318036810 taught by Professor Waung during the Spring '09 term at UCLA.

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6b-e-potential - Electrostatic Potential: There are three...

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