Test Index

Electrostatics

Section: Physics

Question : 12 of 32

Marks: +1, -0

Solution:

\;U_{\;\text{initial }}=U_{12}+U_{13}+U_{23}

\;=\;\frac{k q^2}{2 a}-\;\frac{k q^2}{a}-\;\frac{k q^2}{a}=\;\frac{k q^2}{2 a}-\;\frac{2 k q^2}{a}=\;\frac{-3 k q^2}{2 a}

After exchanging the final charge with one of the negative charge, there are two -

q

charges separated by a distance

a

and one

+q

charge separated by a distance '

a

' from one of the

-q

charges.

\; \therefore U_{\;\text{final }}=U_{12}^{'}+U_{13}^{'}+U_{23}^{'}

\;=\;\frac{k q^2}{a}-\;\frac{k q^2}{2 a}-\;\frac{k q^2}{a}=\;\frac{-k q^2}{2 a}

\; \therefore \Delta U=U_{\;\text{final }}-U_{\;\text{initial }}

\; \;\; =-\;\frac{k q^2}{2 a}- \left( \;\frac{-3 k q^2}{2 a} \right) =\;\frac{2 k q^2}{2 a}

\; \;\; =\;\frac{k q^2}{a}= \frac{q^2}{4 \pi \varepsilon_0 a} \;\; \left[ \because k=\frac{1}{4 \pi \varepsilon_0} \right]

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