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Electrostatics

Section: Physics

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Question : 10 of 32

Marks: +1, -0

Three particles of each charge

q

are placed at the vertices of an equilateral triangle of side

L

. The work to be done to decrease the side of the triangle to

\;\frac{L}{2}

[AP EAPCET 21 May 2025 S2]

$\;\frac{1}{4\pi\epsilon_0} \;\frac{q^2}{L}$
$\;\frac{1}{4\pi\epsilon_0} \;\frac{2q^2}{L}$
$\;\frac{1}{4\pi\epsilon_0} \;\frac{3q^2}{L}$
$\;\frac{1}{4\pi\epsilon_0} \;\frac{3q^2}{2L}$

Solution:

Initial potential energy of the system of

U_i=3 \times \;\frac{1}{4\pi\epsilon_0} \cdot \;\frac{q^2}{L}

Final potential energy of the system

U_f=3 \times \;\frac{1}{4\pi\epsilon_0} \cdot \;\frac{q^2}{\frac{L}{2}}=6 \times \;\frac{1}{4\pi\epsilon_0} \cdot \;\frac{q^2}{L}

\therefore

Required work done

W\;=U_f-U_i

\;=6 \times \;\frac{1}{4\pi\epsilon_0} \cdot \;\frac{q^2}{L}-3 \times \;\frac{1}{4\pi\epsilon_0} \cdot \;\frac{q^2}{L}

\;=\;\frac{1}{4\pi\epsilon_0} \;\frac{3q^2}{L}

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