Question

A metallic sphere of radius $1\, m$ is given a negative charge of $20 \times 10^{3} \mu C$ in air. If the bulk modulus is $\left(\frac{1}{4 \pi^{2}}\right) \times 10^{11} N / m ^{2}$, volume strain in the sphere is _______$\times 10^{-5}$

Answer 1

Bulk Modulus $K=\frac{\text { Volume stress }}{\text { Volume strain }}$
Electrostatic pressure $P=\frac{\sigma^{2}}{2 \varepsilon_{0} k}$
$\therefore K =\frac{\sigma^{2}}{2 k \varepsilon_{0}(\text { volume strain })}$
$\therefore $ Volume strain $=\frac{\sigma^{2}}{2 k \varepsilon_{0}} \times \frac{1}{ K }$
$=\left(\frac{ Q }{4 \pi R ^{2}}\right)^{2} \times \frac{1}{2 k \varepsilon_{0}} \times \frac{1}{ K }$
$\ldots .\left(\because \sigma=\frac{ Q }{ A }\right)$
$=\frac{Q^{2}}{16 \pi^{2} \times R^{4} \times 2 \times \varepsilon_{0} \times \frac{1}{4 \pi^{2}} \times 10^{11}}$
$\ldots .($ For air $k =1)$
$=\frac{\left(20 \times 10^{-3}\right)^{2}}{8 \times 1^{4} \times 8.85 \times 10^{-12} \times 10^{11}}$
$=\frac{400 \times 10^{-6}}{8 \times 8.85 \times 10^{-1}}$
$=\frac{50}{8.85} \times 10^{-5}$
$=5.649 \times 10^{-5}$
$\therefore \text { Volume strain }=5.65 \times 10^{-5}$
....(Rounding off to $2$ decimal places)