Question:

Two concentric hollow spherical shells have radii $ r $ and $ R \left(R\,>>\,r\right) $ . A charge $ Q $ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is

Updated On: May 21, 2024
  • $\frac{Q(R+r)}{4 \pi \varepsilon_{0}\left(R^{2}+r^{2}\right)}$
  • $\frac{Q\left(R^{2}+r^{2}\right)}{4 \pi \varepsilon_{0}(R+r)}$
  • $ \frac{Q}{\left(R+r\right)} $
  • zero
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The Correct Option is A

Solution and Explanation

We know that, the surface charge densities is $\sigma=\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)}\,\,\,...(i)$ The electric potential at the centre $V =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\sigma \times 4 \pi r^{2}}{r}+\frac{\sigma \times 4 \pi R^{2}}{R}\right] $ $=\frac{1}{4 \pi \varepsilon_{0}} \times \sigma \times 4 \pi\left[\frac{r^{2}}{r}+\frac{R^{2}}{R}\right]$ $=\frac{\sigma}{\varepsilon_{0}}(r+R) $ and $ V=\frac{Q}{\varepsilon_{0} \cdot 4 \pi\left(r^{2}+R^{2}\right)} \cdot(r+R) $ $\Rightarrow V =\frac{Q}{4 \pi \varepsilon_{0}} \cdot \frac{(r+R)}{\left(r^{2}+R^{2}\right)}$
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Concepts Used:

Gauss Law

Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.

Gauss Law:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

Gauss Law Formula:

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.