Question

A spherical conducting shell has an inner radius of 0.2 m and an outer radius of 0.4 m. There is no charge on the outer surface but the shell does have a nonzero net charge. All charges remain at rest. The potential at the outer surface is -2 V. The potential at the center of the cavity is g

Answer 1

Answer:

$P_c=-4v$

Explanation:

From the question we are told that:

Inner radius $r_1=0.2$

Outer radius$r_2=0.4$

Potential at the outer surface is $P_o= -2 V$

Generally the equation for Potential at the outer surface P_o  is mathematically given by

$P_o=\frac{KQ}{r_2}$

$-2=\frac{KQ}{0.4}$

Therefore

$KQ=-0.8v$

Generally the equation for Potential at Center of outer cavity $P_c$  is mathematically given by

$P_c=\frac{KQ}{r_1}$

$P_c=\frac{-0.8}{0.2}$

$P_c=-4v$

Therefore the potential at the center of cavity

$P_c=-4v$

Answer 2

The potential at the center of the cavity of the spherical conducting shell is -0.4 V.

What is electric potential of spherical shell?

Electric potential of spherical shell is the total amount of work required to move a charge in s spherical shell from one point to other.

Electric potential of spherical shell at outer surface can be given as,

$P_{outer}=k\dfrac{Q}{r_o}$

Electric potential of spherical shell at center of outer gravity can be given as,

$P_{cavity}=k\dfrac{Q}{r_i}$

Here, (Q) is the charge, (k) is the constant and $(r_o, r_i)$ is the outer and inner radius.

A spherical conducting shell has an outer radius of 0.4 m and the potential at the outer surface is -2 V. Thus, the potential at the outer surface can be find out using the above formula as,

$-2=k\dfrac{Q}{0.4}\\kQ=-0.8\rm V$

The spherical conducting shell has an inner radius of 0.2 m. Thus, the potential at the outer surface can be find out using the above formula as,

$P_{cavity}=k\dfrac{Q}{0.2}\\P_{cavity}=\dfrac{-0.8}{0.2}\\P_{cavity}=-0.4\rm V$

Thus, the potential at the center of the cavity of the spherical conducting shell is -0.4 V.

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