Four electric charges +q, +q, -q, and -q are | vedclass

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Question

(A) Let the square be $ABCD$ with side length $a = 2r$. The charges are placed at corners: $A(+q), B(+q), C(-q), D(-q)$.Let the point $P$ be the midpo

Vedclass · Accepted Answer

$\frac{1}{4\pi\epsilon_0} \frac{2q}{r} [\frac{1}{\sqrt{5}} - 1]$

Vedclass · Answer

(A) Let the square be $ABCD$ with side length $a = 2r$. The charges are placed at corners: $A(+q), B(+q), C(-q), D(-q)$.Let the point $P$ be the midpoint between the two negative charges at $C$ and $D$.The distance from $P$ to $C$ is $r$ and from $P$ to $D$ is $r$.The distance from $P$ to $A$ is $\sqrt{(2r)^2 + r^2} = \sqrt{5r^2} = r\sqrt{5}$.The distance from $P$ to $B$ is $\sqrt{(2r)^2 + r^2} = r\sqrt{5}$.The electric potential $V$ at point $P$ is the sum of potentials due to all four charges: $V = \frac{1}{4\pi\epsilon_0} [\frac{q}{r_A} + \frac{q}{r_B} + \frac{-q}{r_C} + \frac{-q}{r_D}]$.Substituting the distances: $V = \frac{1}{4\pi\epsilon_0} [\frac{q}{r\sqrt{5}} + \frac{q}{r\sqrt{5}} - \frac{q}{r} - \frac{q}{r}]$.$V = \frac{1}{4\pi\epsilon_0} [\frac{2q}{r\sqrt{5}} - \frac{2q}{r}]$.$V = \frac{1}{4\pi\epsilon_0} \frac{2q}{r} [\frac{1}{\sqrt{5}} - 1]$.

Four electric charges $+q, +q, -q$, and $-q$ are placed in order at the corners of a square of side $2r$. The electric potential at a point midway between the two negative charges is

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