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A solid sphere of radius R is concentric with a conducting spherical shell that carries charge +qshell and has an inner radius 2R and an outer radius 3R. If the electrostatic potential at the common center of the sphere and shell is the same as the potential at infinity, what is the charge q sphere on the solid sphere if a.) the sphere is conducting and b.) the sphere is nonconducting and has a uniform charge distribution throughout it?

Respuesta :

Chrisnando

Answer:

a) [tex]Q_1 = \frac{-6Q}{5}[/tex]

b) [tex]Q_1 = -\frac{Q}{4}[/tex]

Explanation:

Given:

Radius = R

Inner radius = 2R

Outer radius = 3R

a) For charge when sphere is conducting:

Lets take Q1 as the charge on inner sphere, which means a charge of

-Q1 wil be induced on the outer sphere.

Let's find the net potential at center of the sphere,

We now use the expression:

[tex]\frac{kQ_1}{R}-\frac{kQ_2}{2R}+\frac{k(Q+Q_1)}{3R} = 0 [/tex]

Multiplying through by R, we have:

[tex] = Q_1 -\frac{Q_1}{2}+\frac{Q+Q_1}{3} = 0[/tex]

[tex]Q_1 = \frac{-6Q}{5}[/tex]

b) for charge when sphere is non-conducting.

Let thee net field inside of the sphere beween R1<r<R2 be considered as zero.

Therefore the inner surface charge induced in the sphere will be -Q as to get net charge close to zero.

For zero net potential at the center, we have:

[tex] \frac{3kQ_1}{2R}-\frac{KQ_1}{2R}+\frac{k(Q+Q_1)}{3R}=0[/tex]

[tex] 4kQ_1+kQ = 0[/tex]

[tex]Q_1 = -\frac{Q}{4}[/tex]