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A conducting sphere of radius $10 \;cm$ has an unknown charge. If the electric field $20\; cm$ from the centre of the sphere is $1.5 \times 10^{3} \;N / C$ and points radially inward, what is the net charge (in $n\;C$) on the sphere?
$3.33$
$6.67$
$8.97$
$11.56$
Solution
Electric field intensity $(E)$ at a distance $(d)$ from the centre of a sphere containing net charge $q$ is given by the relation,
$E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d^{2}}$
Where, $q=$ Net charge $=1.5 \times 10^{3} \,N / C$
$d =$ Distance from the centre $=20\, cm =0.2 \,m$
$\varepsilon_{0}=$ Permittivity of free space and $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,Nm ^{2} \,C ^{-2}$
Therefore,
$=6.67 \times 10^{9}\, C =6.67 \,n\,C$
$q=E\left(4 \pi \varepsilon_{0}\right) d^{2}=\frac{1.5 \times 10^{3}}{9 \times 10^{9}}$
Therefore, the net charge on the sphere is $6.67 \,n\,C .$
