Question:
A two point charges $4 q$ and $-q$ are fixed on the $x$-axis at
$x=-\frac{d}{2}$ and $x=\frac{d}{2}$, respectively. If a third point charge
' $q$ ' is taken from the origin to $x=d$ along the semicircle as shown in the figure, the energy of the charge will:
Correct Option: , 4
Solution:
(4) Change in potential energy, $\Delta u=q\left(V_{f}-V_{i}\right)$
Potential of $-q$ is same as initial and final point of the path.
$\Delta u=q\left(\frac{k 4 q}{3 d / 2}-\frac{k 4 q}{d / 2}\right)=-\frac{4 q^{2}}{3 \pi \varepsilon_{0} d}$
-ve sign shows the energy of the charge is decreasing.