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Question

Two equal charges $$q$$ are placed at a distance $$2a$$ and a third charge $$-2q$$ is placed at the midpoint. The potential energy of the system is 


A
9q28πϵ0a
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B
q28πϵ0a
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C
7q28πϵ0a
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D
6q28πϵ0a
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Solution

The correct option is C $$\dfrac{-7q^2}{8\pi\epsilon_0a}$$
Given two equal charges $$'q'$$ are at a distance $$2a$$.
A third charge $$-2q$$ is placed at the mid-point. We have to find the potential energy of the system.

Let $$O$$ be the midpoint of the charges.
So, potential energy of the system is

$$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{(-2q)(q)}{a}-\dfrac{(2q)(q)}{a}+\dfrac{(q)(q)}{2a} \right ]$$
$$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-2q^{2}}{a}-\dfrac{2q^{2}}{a}+\dfrac{q^{2}}{2a} \right ]$$
$$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-4q^{2}}{a}+\dfrac{q^{2}}{2a} \right ]$$
$$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-8q^{2}a+q^{2}a}{2a^{2}} \right ]$$
$$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-7q^{2}a}{2a^{2}} \right ]$$
$$=\dfrac{1}{4\pi \varepsilon _{0}}\left [ \dfrac{-7q^{2}}{2a} \right ]$$
$$=\dfrac{-7q^{2}}{8\pi \varepsilon _{0}a}$$


Physics

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