Electrostatics 7 Question 33
35. Assume the earth to be a sphere of uniform mass density. Calculate this energy, given the product of the mass and the radius of the earth to be $2.5 \times 10^{31} kg-m$.
(a) If the same charge of $Q$ as in part (a) above is given to a spherical conductor of the same radius $R$, what will be the energy of the system?
(1992, 10M)
(b) A charge of $Q$ is uniformly distributed over a spherical volume of radius $R$. Obtain an expression for the energy of the system.
(c) What will be the corresponding expression for the energy needed to completely disassemble the planet earth against the gravitational pull amongst its constituent particles?
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Solution:
- (a) In this case the electric field exists from centre of the sphere to infinity. Potential energy is stored in electric field, with energy density
$$ u=\frac{1}{2} \varepsilon _0 E^{2} \quad(\because \text { Energy } / \text { Volume }) $$
(i) Energy stored within the sphere $\left(U _1\right)$
Electric field at a distance $r$ is
$$ \begin{aligned} & E=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q}{R^{3}} \cdot r \\ & u=\frac{1}{2} \varepsilon _0 E^{2}=\frac{\varepsilon _0}{2} \quad \frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q}{R^{3}} r^{2} \end{aligned} $$
Volume of element,
$$ d V=\left(4 \pi r^{2}\right) d r $$
Energy stored in this volume, $d U=u(d V)$
$$ \begin{aligned} d U & =\left(4 \pi r^{2} d r\right) \frac{\varepsilon _0}{2} \frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q^{3}}{R^{2}} \\ d U & =\frac{1}{8 \pi \varepsilon _0} \cdot \frac{Q^{2}}{R^{6}} \cdot r^{4} d r \\ \therefore \quad U _1 & =\int _0^{R} d U=\frac{1}{8 \pi \varepsilon _0} \frac{Q^{2}}{R^{6}} \int _0^{R} r^{4} d r \\ & =\frac{Q^{2}}{40 \pi \varepsilon _0 R^{6}}\left[r^{5}\right] _0^{R} \\ U _1 & =\frac{1}{40 \pi \varepsilon _0} \cdot \frac{Q^{2}}{R} \end{aligned} $$
(ii) Energy stored outside the sphere $\left(U _2\right)$
Electric field at a distance $r$ is
$$ \begin{aligned} E & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q}{r^{2}} \\ \therefore \quad u & =\frac{1}{2} \varepsilon _0 E^{2}=\frac{\varepsilon _0}{2} \quad \frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q^{2}}{r^{2}} \end{aligned} $$
$$ \begin{aligned} d U & =\left(4 \pi r^{2} d r\right) \\ d U & =u \cdot d V \\ & =\left(4 \pi r^{2} d r\right) \frac{\varepsilon _0}{2} \frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q^{2}}{r^{2}} \\ d U & =\frac{Q^{2}}{8 \pi \varepsilon _0} \frac{d r}{r^{2}} \\ \therefore \quad U _2 & =\int _R^{\alpha} d U=\frac{Q^{2}}{8 \pi \varepsilon _0} \cdot \int _R^{\alpha} \frac{d r}{r^{2}} \\ U _2 & =\frac{Q^{2}}{8 \pi \varepsilon _0 R} \end{aligned} $$
Therefore, total energy of the system is
$$ \begin{aligned} U & =U _1+U _2=\frac{Q^{2}}{40 \pi \varepsilon _0 R}+\frac{Q^{2}}{8 \pi \varepsilon _0 R} \\ \text { or } \quad U & =\frac{3}{20} \frac{Q^{2}}{\pi \varepsilon _0 R} \end{aligned} $$
(b) Comparing this with gravitational forces, the gravitational potential energy of earth will be
$$ U=-\frac{3}{5} \frac{G M^{2}}{R} $$
$$ \begin{aligned} & \text { by replacing } Q^{2} \text { by } M^{2} \text { and } \frac{1}{4 \pi \varepsilon _0} \text { by } G \\ & g=\frac{G M}{R^{2}} \\ & \therefore \quad G=\frac{g R^{2}}{M} \\ & U=\frac{-3}{5} M g R \end{aligned} $$
Therefore, energy needed to completely disassemble the earth against gravitational pull amongst its constituent particles will be given by
$$ E=|U|=\frac{3}{5} M g R $$
Substituting the values, we get
$$ \begin{aligned} & E=\frac{3}{5}\left(10 m / s^{2}\right)\left(2.5 \times 10^{31} kg-m\right) \\ & E=1.5 \times 10^{32} J \end{aligned} $$
(c) This is the case of a charged spherical conductor of radius $R$, energy of which is given by $=\frac{1}{2} \frac{Q^{2}}{C}$
or
$$ U=\frac{1}{2} \cdot \frac{Q^{2}}{4 \pi \varepsilon _0 R} \text { or } U=\frac{Q^{2}}{8 \pi \varepsilon _0 R} $$
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