Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge –Q from its centre to infinity is

\frac{\sqrt{2}Q^2}{4\pi\epsilon_0 a}

\frac{\sqrt{2}Q^2}{\pi\epsilon_0 a}

\frac{Q^2}{2\pi\epsilon_0 a}

Identify Geometry: The distance $r$ from any corner of the square (side $a$ ) to the center is half the diagonal. $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$ .
Calculate Potential at Center ( $V_c$ ): The electric potential is a scalar quantity, so we sum the potentials due to the four charges $+Q$ at the corners [Source 157]. $V_c = 4 \times \frac{1}{4\pi\epsilon_0} \frac{Q}{r} = \frac{1}{\pi\epsilon_0} \frac{Q}{(a/\sqrt{2})} = \frac{\sqrt{2}Q}{\pi\epsilon_0 a}$
Calculate Potential Energy ( $U$ ): The potential energy of the charge $-Q$ at the center is $U = (-Q)V_c$ [Source 36, 161]. $U_{center} = -Q \left( \frac{\sqrt{2}Q}{\pi\epsilon_0 a} \right) = -\frac{\sqrt{2}Q^2}{\pi\epsilon_0 a}$
Calculate Work Done ( $W$ ): The work done to remove the charge to infinity (where potential energy is zero) is the change in potential energy. $W = U_{final} - U_{initial} = 0 - U_{center} = \frac{\sqrt{2}Q^2}{\pi\epsilon_0 a}$

Practice Mode Available

Join thousands of students and practice with AI-generated mock tests.