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

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(B) The potential $V$ at the center of a square due to four identical charges $q$ at the corners is given by $V = 4 	imes \frac{kq}{r}$,where $r$ is

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$1500 \sqrt{2} 	ext{ V}$

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(B) The potential $V$ at the center of a square due to four identical charges $q$ at the corners is given by $V = 4 	imes \frac{kq}{r}$,where $r$ is the distance from the center to any corner.For a square of side $a = 8 	ext{ cm} = 0.08 	ext{ m}$,the diagonal length is $a\sqrt{2}$. The distance $r$ from the center to a corner is half the diagonal: $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Given $q = \frac{10}{3} 	imes 10^{-9} 	ext{ C}$,$a = 8 	imes 10^{-2} 	ext{ m}$,and $k = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$.$V = 4 	imes \frac{k 	imes q}{a / \sqrt{2}} = \frac{4 	imes 9 	imes 10^9 	imes (\frac{10}{3} 	imes 10^{-9}) 	imes \sqrt{2}}{8 	imes 10^{-2}}$.$V = \frac{4 	imes 9 	imes \frac{10}{3} 	imes \sqrt{2}}{8 	imes 10^{-2}} = \frac{120 \sqrt{2}}{8 	imes 10^{-2}} = 15 \sqrt{2} 	imes 100 = 1500 \sqrt{2} 	ext{ V}$.

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

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