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Consider a spherical surface of radius 4m centred at the origin . Point +q and -2q are fixed at points $\;A (2m , 0 , 0)\;$ and $\;B (8m , 0 ,0)\;$ respectively . The potential on the surface of sphere is


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Answer : (d) zero
Explanation :
Let $\;P (x , y ,z)\;$ be any point on the sphere . From the property of the sphere
Further ,
and $\;PB= \sqrt{(x-8)^2+y^2+z^2}$
$=2 \sqrt{20-4x}$
Let $\;V_{P}\;$ be potential at P
$V_{P}=\large\frac{1}{4 \pi \in_{0}}\;[\large\frac{q}{PA}-\large\frac{2\;q}{PB}]$
$=\large\frac{1}{4 \pi \in_{0}}\;[\large\frac{q}{\sqrt{20-4x}}-\large\frac{2\;q}{2 \sqrt{20-4x}}]$
answered Feb 12, 2014 by yamini.v

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