Question

A spherical shell of radius \[R = 1.5\,{\text{cm}}\] has a charge \[q = 20\,\mu {\text{C}}\] uniformly distributed over it. The force exerted by one half over the other half is
A. zero
B. \[{10^{ - 2}}\,{\text{N}}\]
C. \[500\,{\text{N}}\]
D. \[2000\,{\text{N}}\]

Answer 1

Hint:Use the equation for the electrostatic pressure exerted on the surface of the half portion of the shell. Use the value of surface charge density in this equation. Substitute the value of this derived equation in the equation for pressure to determine the force exerted by the half portion on the other half portion of the spherical shell.

Formulae used:
The electrostatic pressure \[P\] exerted by charges on a surface is
\[\Rightarrow P = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\] …… (1)
Here, \[\sigma \] is the surface charge density and \[{\varepsilon _0}\] is the permittivity of free space.
The surface charge density \[\sigma \] for a spherical surface is
\[\Rightarrow\sigma = \dfrac{q}{{4\pi {R^2}}}\] …… (2)
Here, \[q\] is the charge distributed over the surface and \[R\] is the radius of the sphere.
The pressure \[P\] on an object is given by
\[\Rightarrow P = \dfrac{F}{A}\] …… (3)
Here, \[F\] is the force on the object and \[A\] is the area on which the force is applied.

Complete step by step solution:
We have given that the radius of the spherical shell is \[1.5\,{\text{cm}}\] and the charge which is uniformly distributed over the surface of shell is \[20\,\mu {\text{C}}\].
\[\Rightarrow R = 1.5\,{\text{cm}}\]
\[\Rightarrow q = 20\,\mu {\text{C}}\]
Convert the unit of radius in the SI system of units.
\[\Rightarrow R = \left( {1.5\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)\]
\[ \Rightarrow R = 1.5 \times {10^{ - 2}}\,{\text{m}}\]
Convert the unit of the charge in the SI system of units.
\[\Rightarrow q = \left( {20\,\mu {\text{C}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{\text{C}}}}{{1\,\mu {\text{C}}}}} \right)\]
\[ \Rightarrow q = 20 \times {10^{ - 6}}\,{\text{C}}\]
Hence, the charge distributed over the spherical shell is \[20 \times {10^{ - 6}}\,{\text{C}}\].
The force exerted by one half on the other half of the spherical shell is the force exerted by the inner half circular surface on the opposite half circular surface of the spherical shell.
Hence, the force is exerted on the circular area (\[\pi {R^2}\]) of the shell.
Substitute \[\dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\] for \[P\] and \[\pi {R^2}\] for \[A\] in equation (3).
\[\Rightarrow\dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}} = \dfrac{F}{{\pi {R^2}}}\]
Rearrange the above equation for \[F\].
\[ \Rightarrow F = \dfrac{{\pi {R^2}{\sigma ^2}}}{{2{\varepsilon _0}}}\]
Substitute \[\dfrac{q}{{4\pi {R^2}}}\] for \[\sigma \] in the above equation.
\[ \Rightarrow F = \dfrac{{\pi {R^2}{{\left( {\dfrac{q}{{4\pi {R^2}}}} \right)}^2}}}{{2{\varepsilon _0}}}\]
\[ \Rightarrow F = \dfrac{{{q^2}}}{{32\pi {R^2}{\varepsilon _0}}}\]
Substitute \[20 \times {10^{ - 6}}\,{\text{C}}\] for \[q\], \[3.14\] for \[\pi \], \[1.5 \times {10^{ - 2}}\,{\text{m}}\] for \[R\] and \[8.85 \times {10^{ - 12}}\,{\text{F}} \cdot {{\text{m}}^{{\text{ - 1}}}}\] for \[{\varepsilon _0}\] in the above equation.
\[ \Rightarrow F = \dfrac{{{{\left( {20 \times {{10}^{ - 6}}\,{\text{C}}} \right)}^2}}}{{32\left( {3.14} \right){{\left( {1.5 \times {{10}^{ - 2}}\,{\text{m}}} \right)}^2}\left( {8.85 \times {{10}^{ - 12}}\,{\text{F}} \cdot {{\text{m}}^{{\text{ - 1}}}}} \right)}}\]
\[ \therefore F = 2000\,{\text{N}}\]
Therefore, the force exerted by the one half on the other half is \[2000\,{\text{N}}\].

Hence, the correct option is D.

Note:The students may think that force exerted on one half portion of the spherical shell on the other half portion is on the whole surface area of the half spherical shell. But the actual pressure is exerted on the area of the inner circular portion of the sphere only. Hence, while substituting in the pressure equation (3), the area should be substituted by the area of the circle on which the force is to be applied.