Question

Three concentric metal shells $A,B$ and $C$ of respective radii $a,b$ and $c(a<b<c)$ have surface charge densities $+\sigma ,-\sigma$ and $+\sigma$ respectively. The potential of shell $B$ is:

• A. $\frac{\sigma }{{ϵ}_0}\left[\frac{b^2-c^2}{b}+a\right]$
• B. $\frac{\sigma }{{ϵ}_0}\left[\frac{b^2-c^2}{c}+a\right]$
• C. $\frac{\sigma }{{ϵ}_0}\left[\frac{a^2-b^2}{a}+c\right]$
• D. $\frac{\sigma }{{ϵ}_0}\left[\frac{a^2-b^2}{b}+c\right]$

Answer 1

The correct option is D

$\frac{\sigma }{{ϵ}_0}\left[\frac{a^2-b^2}{b}+c\right]$

Step 1: Given terms:

Surface charge densities on the shell $A$ with radius $a=+\sigma$

Surface charge densities on the shell $B$ with radius $b=-\sigma$

Surface charge densities on the shell $C$ with radius$c=+\sigma$

Step 2: Formula used

Surface charge density $\sigma =\frac{q}{Area}$

Electric potential $V_{inside}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r}$

Electric potential $V_{Outside}=V_{surface}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{R}$

Where $q$ is the charge $r$ is the distance of the point at which the potential is calculated from the charged surface and ${\epsilon }_0$ is the permeability in free space and $R$ is the distance of the charge from the surface.

Step 3: Calculating potential

Surface Area of a Sphere$=4{\mathrm{\pi r}}^2$, where $r$ is the radius of the circle.

$$\begin{gathered} \because \sigma =\frac{q}{Area} \\ \Rightarrow q=\sigma (Area) \end{gathered}$$

$\therefore q_A=+\sigma 4\pi a^2$

$q_B=-\sigma 4\pi b^2$

$q_C=+\sigma 4\pi c^2$

Again, we know formula of electric potential for charge $q$, $V=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r}$

The electric potential at $B$ is equal to the sum of all the electric potential as the three surfaces $A,B$ and $C$

$V_B=V_A+V_B+V_C$, $B$ is outside point of $A$ and $B$is the inside point of $C$ and $B$ is on the surface of $B$.

$\therefore V_B=V_{BA}+V_{BB}+V_{BC}$

Where $V_{BA}=$Potential at $B$ due to $A.$

$V_{BB}=$Potential at $B$ due to $B.$

$V_{BC}=$Potential at $B$ due to $C.$

The distance of potential $V_{BA}$ will be $b.$ as $\left(a=b\right)$ for $V_{BA}.$

Thus, $\Rightarrow V_B=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_A}{b}+\frac{q_B}{b}+\frac{q_C}{c}\right)$

$V_A=V_{outside}$

$V_B=$$V_{Surface}$

$V_C=V_{inside}$

Substituting the value of $q_A,q_B$ and $q_C$ we have

$$\begin{gathered} \therefore V_B=\frac{1}{4\pi {\epsilon }_0}\left(\frac{4\pi a^2\sigma }{b}-\frac{4\pi b^2\sigma }{b}+\frac{4\pi c^2\sigma }{c}\right) \\ =\frac{\sigma }{{\epsilon }_0}\left[\frac{a^2-b^2}{b}+c\right] \end{gathered}$$

Hence, the correct option is option D.