- Home
- Standard 12
- Physics
1. Electric Charges and Fields
medium
Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. A third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now
A
$-2 E$
B
$-E$
C
$0$
D
$E$
(KVPY-2014)
Solution
(b)
Electrostatic energy of two equal charges of magnitude $Q$ placed $d$ distance apart is
$E=\frac{k q_1 q_2}{r_{12}}=\frac{k Q^2}{d} \quad \dots(i)$
Now, when a third charge $\left(-\frac{Q}{2}\right)$ is placed at mid-point of these charges, then electrostatic energy of system is
$E^{\prime}=\frac{k q_1 q_2}{r_{12}}+\frac{k q_2 q_3}{r_{23}}+\frac{k q_1 q_3}{r_{13}}$
$=\frac{k Q^2}{d}-\frac{k Q^2 / 2}{d / 2}-\frac{k Q^2 / 2}{d / 2}$
$=-\frac{k Q^2}{d}=-E$
[from Eq. $(i)$]
Standard 12
Physics
