Question

There is a -ve charge -$q$ as shown in the figure. Take the potential to be zero at infinity. If $A$ and $B$ are two points in the charge's vicinity, then the potential at-

• A. $A$ is less than at $B$
• B. $A$ is more than that at $B$
• C. $A$ is greater in magnitude than at $B$
• D. $A$ is smaller in magnitude than at $B$

Answer 1

Answer: (A), (C)

Potential at Point $A$ will be given by,

$V_A=\frac{K(-q)}{A}=\frac{-Kq}{A}$

Potential at Point $B$ will be given by,

$V_B=\frac{K(-q)}{B}=\frac{-Kq}{B}$

Now, As $A<B$,

So, the potential at $A$ is greater magnitude than at $B$.

But, as due to negative sign convention potential is greater at $B$ than at $A$.