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

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Question

(B) Let the square be $ABCD$ with side length $2L$. The charges are placed as follows: $A(+q), B(+q), C(-q), D(-q)$. Point $P$ is the midpoint of side

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{2q}{L} \left(1 - \frac{1}{\sqrt{5}}\right)$

Vedclass · Answer

(B) Let the square be $ABCD$ with side length $2L$. The charges are placed as follows: $A(+q), B(+q), C(-q), D(-q)$. Point $P$ is the midpoint of side $AB$.The distances from point $P$ to the charges are:$AP = L$$BP = L$$DP = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$$CP = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$The total electric potential $V$ at point $P$ is the sum of potentials due to individual charges:$V = V_A + V_B + V_C + V_D$$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{CP} + \frac{-q}{DP} \right)$$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{L} + \frac{q}{L} - \frac{q}{L\sqrt{5}} - \frac{q}{L\sqrt{5}} \right)$$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{2q}{L} - \frac{2q}{L\sqrt{5}} \right)$$V = \frac{1}{4 \pi \varepsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$

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 $P$,which is midway between the two positive charges,is:

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