Charges of + frac{10}{3} times 10^{-9} , C ar | vedclass

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(B) The potential $V$ at the center $O$ of the square due to four identical charges $q$ at the corners is given by the sum of potentials due to each c

Vedclass · Accepted Answer

$1500\sqrt{2} \, V$

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(B) The potential $V$ at the center $O$ of the square due to four identical charges $q$ at the corners is given by the sum of potentials due to each charge.$V = 4 	imes \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r}$,where $r$ is the distance from each corner to the center.For a square of side $a = 8 \, cm = 8 	imes 10^{-2} \, m$,the distance $r$ from the corner to the center is half the diagonal length:$r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} = \frac{8 	imes 10^{-2}}{\sqrt{2}} \, m$.Given $q = \frac{10}{3} 	imes 10^{-9} \, C$ and $\frac{1}{4\pi \varepsilon_0} = 9 	imes 10^9 \, N \cdot m^2/C^2$.Substituting the values:$V = 4 	imes (9 	imes 10^9) 	imes \frac{\frac{10}{3} 	imes 10^{-9}}{\frac{8 	imes 10^{-2}}{\sqrt{2}}}$$V = 4 	imes 9 	imes 10^9 	imes \frac{10}{3} 	imes 10^{-9} 	imes \frac{\sqrt{2}}{8 	imes 10^{-2}}$$V = 120 	imes \frac{\sqrt{2}}{8 	imes 10^{-2}} = 15 	imes 10^2 	imes \sqrt{2} = 1500\sqrt{2} \, V$.

Charges of $+ \frac{10}{3} \times 10^{-9} \, C$ are placed at each of the four corners of a square of side $8 \, cm$. The potential at the intersection of the diagonals is

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