A particle carrying charge +q is placed at the center of a thick-walled conducting shell that has inner radius R and outer radius 2R and carries charge −2q. A thin-walled conducting shell of radius 5R carries charge +2q and is concentric with the thick-walled shell. Define V = 0 at infinity. Calculate all distances from the particle at which the electrostatic potential is zero.

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kollybaba55 kollybaba55 · Answer 1 · 2020-04-13T06:24:35+02:00

Answer:

10R/11 and 5R/2

Explanation:

The radius of the conducting shell = R,

Electrostatic potential inside the shell (r<R) = kq/R

Electrostatic potential outside the shell (r>R) = kq/r

If x is the point of zero potential

Electrostatic potential for inner shell, [tex]V_{1} = \frac{kq}{X - R}[/tex]

Electrostatic potential for outer shell, [tex]V_{2} = \frac{-2kq}{X - 2R}[/tex]

Electrostatic potential for the thin walled shell, [tex]V_{3} = \frac{2kq}{X - 5R}[/tex]

[tex]V_{1} + V_{2} + V_{3} = 0[/tex]

[tex]\frac{kq}{X-R} - \frac{2kq}{X-2R} + \frac{2kq}{X-5R} = 0[/tex]

[tex]\frac{1}{X-R} - \frac{2}{X-2R} + \frac{2}{X-5R} = 0\\(X-R) - 2(X-R)(X-5R)+2(X-R)(X-2R) = 0[/tex]

The values of X=r that satisfy the above equation are 10R/11 and 5R/2