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Question

(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 \d

Vedclass · Accepted Answer

$-E$

Vedclass · Answer

(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. $(i)$]

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

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