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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
Question
Two charges -q each are separated by distance 2d. A third charge +q is kept at mid point O. Find potential energy of +q as a function of small distance x from O due to -q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.
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Answer

$\text{U}=\frac{1}{4\pi\epsilon_0}\left\{\frac{-\text{q}^2}{(\text{d}-\text{x})}+\frac{-\text{q}^2}{(\text{d}+\text{x})}\right\}$
$\Rightarrow\ \text{U}=\frac{-\text{q}^2}{4\pi\epsilon_0}\frac{2\text{q}}{(\text{d}^2-\text{x}^2)}$
$\Rightarrow\ \frac{\text{dU}}{\text{dx}}=\frac{-\text{q}^2 2\text{d}}{4\pi\epsilon_0}\frac{2\text{x}}{(\text{d}^2-\text{x}^2)^2}$
Here, $\frac{\text{dU}}{\text{dx}}=0$ at x = 0
x is an equilibrium point.
$\frac{\text{d}^2\text{U}}{\text{dx}^2}=\bigg(\frac{-2\text{dq}^2}{4\pi\epsilon_0}\bigg)\Bigg[\frac{2}{(\text{d}^2-\text{x}^2)^2}-\frac{8\text{x}^2}{(\text{d}^2-\text{x}^2)^3}\Bigg]$
$=\bigg(\frac{-2\text{dq}^2}{4\pi\epsilon_0}\bigg)\frac{1}{(\text{d}^2-\text{x}^2)^3}\bigg[2(\text{d}^2-\text{x}^2)^2-8\text{x}^2\bigg]$
At, x = 0
$\frac{\text{d}^2\text{U}}{\text{dx}^2}=\bigg(\frac{-2\text{dq}^2}{4\pi\epsilon_0}\bigg)\bigg(\frac{1}{\text{d}^6}\bigg)(2\text{d}^2),\text{ which is}<0$
So, unstable equilibrium.

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