Question

Figure shows three spherical and equipotential surfaces $A$, $B$ and $C$ around a point charge $q$. The potential difference $V_A-V_B=V_B-V_C$. If $t_1$ and $t_2$ be the distances between them, then :

• A. $t_1=t_2$
• B. $t_1>t_2$
• C. $t_1<t_2$
• D. $t_1\le t_2$

Answer 1

The correct option is C $t_1<t_2$

The potential around a point charge is $V\left(r\right)=\frac{q}{4\pi {ϵ}_{0}r}$

Let $a,b,c$ be the radii of $A,B,C$ respectively.

Thus, $t_1=b-a$ and $t_2=c-b$

$V_A-V_B=\frac{q}{4\pi {ϵ}_{0}a}-\frac{q}{4\pi {ϵ}_{0}b}=\frac{q{t}_1}{4\pi {ϵ}_{0}ab}$

$V_B-V_C=\frac{q}{4\pi {ϵ}_{0}b}-\frac{q}{4\pi {ϵ}_{0}c}=\frac{q{t}_2}{4\pi {ϵ}_{0}bc}$

But, $V_A-V_B=V_B-V_C$

$\Rightarrow \frac{V_A-V_B}{V_B-V_C}=1\Rightarrow \frac{t_1}{ab}\times \frac{bc}{t_2}=1\Rightarrow \frac{t_1}{t_2}=\frac{a}{c}$

As $a<c$, we have $t_1<t_2$