Question

A test charge ${}^{\prime }{q}^{\prime }$ is moved without acceleration from $A$ to $C$ along the path from $A$ to $B$ and then from $B$ to $C$ in electric field $E$ as shown in the figure. (i) Calculate the potential difference between $A$ and $C$. (ii) At which point (of the two) is the electric potential more and why?

Answer 1

(i)

Electric field is given by:

$\overrightarrow{E}=E\hat{i}=-\frac{dV}{dx}\hat{i}$

Hence, $V_C-V_A=-\int Edx$

$V_A-V_C=-4E$

(ii)

Electric potential is more at point C as electric field is directed from higher potential region to lower potential region.