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

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(D) Let the square be $ABCD$ with side length $2L$. Charges $+q$ are at $A$ and $B$,and charges $-q$ are at $D$ and $C$. Point $P$ is the midpoint of

Vedclass · Accepted Answer

$\frac{1}{4 \pi \epsilon_0} \frac{2q}{L}\left(1-\frac{1}{\sqrt{5}}\right)$

Vedclass · Answer

(D) Let the square be $ABCD$ with side length $2L$. Charges $+q$ are at $A$ and $B$,and charges $-q$ are at $D$ and $C$. Point $P$ is the midpoint of $AB$.Distance $AP = PB = L$.The distance from $P$ to $D$ and $C$ can be found using the Pythagorean theorem in $	riangle ADP$ and $	riangle BCP$:$PD = PC = \sqrt{(2L)^2 + L^2} = \sqrt{4L^2 + L^2} = \sqrt{5}L$.The electric potential $V$ at point $P$ is the sum of potentials due to all four charges:$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{PD} + \frac{-q}{PC} ight)$$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{L} + \frac{q}{L} - \frac{q}{\sqrt{5}L} - \frac{q}{\sqrt{5}L} ight)$$V = \frac{1}{4 \pi \epsilon_0} \left( \frac{2q}{L} - \frac{2q}{\sqrt{5}L} ight)$$V = \frac{1}{4 \pi \epsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} ight)$

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

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