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

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Question

(b) Potential at the centre $O$, $V = 4 	imes \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{{a/\sqrt 2 }}$
where $Q = \frac{{10}}{3} 	imes {10^{ - 9}}

Vedclass · Accepted Answer

$1500\sqrt 2 \,volt$

Vedclass · Answer

(b) Potential at the centre $O$, $V = 4 	imes \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{{a/\sqrt 2 }}$
where $Q = \frac{{10}}{3} 	imes {10^{ - 9}}C$ and $a = 8\,cm\, = \,\,8 	imes {10^{ - 2}}m$
So $V = 5 	imes 9 	imes {10^9} 	imes \frac{{\frac{{10}}{3} 	imes {{10}^{ - 9}}}}{{\frac{{8 	imes {{10}^{ - 2}}}}{{\sqrt 2 }}}}$$ = 1500\sqrt 2 \,volt$

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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