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Consider a metal sphere of radius $R$ that is cut in two parts along a plane whose minimum distance from the sphere's centre is $h$. Sphere is uniformly charged by a total electric charge $Q$. The minimum force necessary to hold the two parts of the sphere together, is
$\frac{{{Q^2}\left( {{R^2} - {h^2}} \right)}}{{4\pi \,{ \in _0}\,{R^4}}}$
$\frac{{{Q^2}}}{{4\pi \,{ \in _0}\,{R^2}}}$
$\frac{{{Q^2}\left( {R - h} \right)}}{{32\pi \,{ \in _0}\,{R^3}}}$
$\frac{{{Q^2}\left( {{R^2} - {h^2}} \right)}}{{32\pi \,{ \in _0}\,{R^4}}}$
Solution
Total force necessary to hold the two parts is $=$ (Cross $-$ section area ) $\times$ Pressure
$ = \pi \left( {{{\rm{R}}^2} – {{\rm{h}}^2}} \right) \times \frac{{{\sigma ^2}}}{{2{ \in _0}}} = \pi \left( {\frac{{{{\rm{R}}^2} – {{\rm{h}}^2}}}{{2{ \in _0}}}} \right) \times {\left( {\frac{{\rm{Q}}}{{4\pi {{\rm{R}}^2}}}} \right)^2}$
$ = \frac{{\left( {{{\rm{R}}^2} – {{\rm{h}}^2}} \right){{\rm{Q}}^2}}}{{32\pi { \in _0}{{\rm{R}}^4}}}$
