Charges of 1 mu C are placed at each of the f | vedclass

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Question

(A) The length of the diagonal of a square with side $a$ is $a \sqrt{2}$.Given side $a = 2 \sqrt{2} 	ext{ m}$,the diagonal length is $d = (2 \sqrt{2}

Vedclass · Accepted Answer

$18 	imes 10^3 	ext{ V}$

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(A) The length of the diagonal of a square with side $a$ is $a \sqrt{2}$.Given side $a = 2 \sqrt{2} 	ext{ m}$,the diagonal length is $d = (2 \sqrt{2}) 	imes \sqrt{2} = 4 	ext{ m}$.The distance $r$ of each vertex from the center (intersection of diagonals) is half the diagonal length:$r = \frac{d}{2} = \frac{4}{2} = 2 	ext{ m}$.The electric potential $V$ at the center due to four identical charges $q = 1 \mu C = 10^{-6} 	ext{ C}$ at distance $r$ is given by:$V = 4 	imes \frac{k q}{r}$,where $k = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$.Substituting the values:$V = 4 	imes \frac{9 	imes 10^9 	imes 10^{-6}}{2}$$V = 2 	imes 9 	imes 10^3 = 18 	imes 10^3 	ext{ V}$.

Charges of $1 \mu C$ are placed at each of the four corners of a square of side $2 \sqrt{2} \text{ m}$. The potential at the point of intersection of the diagonals is . . . . . . .

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