Four point charges -Q, -q, 2q and 2Q are plac | vedclass

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Question

(A) Let the distance from the center of the square to each corner be $r$. The electric potential $V$ at the center due to a point charge $q_i$ at dist

Vedclass · Accepted Answer

$Q = -q$

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(A) Let the distance from the center of the square to each corner be $r$. The electric potential $V$ at the center due to a point charge $q_i$ at distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r}$.Since the distance $r$ is the same for all four corners,the total potential at the center is the algebraic sum of the potentials due to each charge:$V_{total} = \frac{1}{4\pi\epsilon_0} \left( \frac{-Q}{r} + \frac{-q}{r} + \frac{2q}{r} + \frac{2Q}{r} \right)$$V_{total} = \frac{1}{4\pi\epsilon_0 r} (-Q - q + 2q + 2Q)$$V_{total} = \frac{1}{4\pi\epsilon_0 r} (Q + q)$For the potential at the center to be zero,we must have $V_{total} = 0$,which implies:$Q + q = 0$$Q = -q$

Four point charges $-Q, -q, 2q$ and $2Q$ are placed,one at each corner of a square. The relation between $Q$ and $q$ for which the potential at the center of the square is zero is:

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