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Q. The electric potential $V$ in volts in a region of space is given by $V=a x^{2}+a y^{2}+2 a z^{2} .$ The work done by field when a $2\, \mu C$ test charge moves from point $(0,0,0.1) m$ to origin is $5 \times 10^{-5} J$. Determine $a$.

Electrostatic Potential and Capacitance

Solution:

$V_{1} =a(0)^{2}+a(0)^{2}+2 a(0.1)^{2}$
$=\frac{a}{50}$
$V_{0} =0 \rightarrow$ at origin.
$W_{e l} =q\left(V_{1}-V_{0}\right)$
$\Rightarrow 5 \times 10^{-5}=2 \times 10^{-6}\left[\frac{a}{50}-0\right]$
$\Rightarrow a =1250$