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

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Question

(C) Let the corners of the square be $A, B, C, D$ in order,with charges $+q$ at $A$ and $B$,and $-q$ at $C$ and $D$. The side length is $2r$. Point $P

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{2q}{r}\left[\frac{1}{\sqrt{5}}-1\right]$

Vedclass · Answer

(C) Let the corners of the square be $A, B, C, D$ in order,with charges $+q$ at $A$ and $B$,and $-q$ at $C$ and $D$. The side length is $2r$. Point $P$ is the midpoint of side $CD$. Thus,the distances are: $DP = PC = r$. The distances from $A$ and $B$ to $P$ are: $AP = BP = \sqrt{AD^2 + DP^2} = \sqrt{(2r)^2 + r^2} = \sqrt{5r^2} = r\sqrt{5}$. The electric potential $V_P$ at point $P$ is the sum of potentials due to all four charges: $V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{CP} + \frac{-q}{DP} \right]$ $V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{r\sqrt{5}} + \frac{q}{r\sqrt{5}} - \frac{q}{r} - \frac{q}{r} \right]$ $V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{2q}{r\sqrt{5}} - \frac{2q}{r} \right]$ $V_P = \frac{1}{4 \pi \varepsilon_0} \frac{2q}{r} \left[ \frac{1}{\sqrt{5}} - 1 \right]$

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 $P$ midway between the two negative charges is:

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