Question

please solve the attached jee question​

Answer 1

Question :–

• A uniformly charged thin spherical shell of radius R carries uniform surface charge density of $\sigma$ per unit area. It is made of two hemispherical shells, hold together by pressing them with force F. F is equal to ?

ANSWER :–

GIVEN :–

• A uniformly charged thin spherical shell of radius R carries uniform surface charge density of $\sigma$ per unit area.

• Sphere is made of two hemispherical shells, hold together by pressing them with force F.

TO FIND :–

• Force = ?

SOLUTION :–

• We know that , Electrostatic pressure at a point on the surface on the uniformly charge sphere is –

$\\ \: \: \: \longrightarrow \: \: \large { \boxed{ \bold{P = \dfrac{ { \sigma}^{2} }{2 \in _{0} } }}}$

• And –

$\\ \: \: \: \longrightarrow \: \: \large { \boxed{ \bold{P = \dfrac{F}{A } }}}$

• So that –

$\\ \: \: \: \implies \: \: { \bold{\dfrac{ { \sigma}^{2} }{2 \in _{0} } = \dfrac{F}{A }}}$

$\\ \: \: \: \implies \: \: { \bold{F= \dfrac{{ \sigma}^{2} }{2 \in _{0} }(A)}}$

$\\ \: \: \: \implies \: \: { \bold{F= \dfrac{{ \sigma}^{2} }{2 \in _{0} }( \pi {R}^{2} )}}$

$\\ \: \: \: \longrightarrow \: \large \: { \boxed{ \bold{F= \dfrac{ \pi{ \sigma}^{2} {R}^{2}}{2 \in _{0} }}}}$

Answer 2

We consider the circular cross section inside the spherical shell (base of hemispheres, in other words) through which flux is executed and whose area is $\sf{A=\pi R^2.}$

As in the fig., two forces are acting on this section. Hence electric flux,

$\longrightarrow\sf{\Phi=2EA\quad\quad\dots(1)}$

By Gauss' Law, we have,

$\longrightarrow\sf{\Phi=\dfrac{q}{\epsilon_0}\quad\quad\dots(2)}$

Comparing (1) and (2),

$\longrightarrow\sf{2EA=\dfrac{q}{\epsilon_0}}$

$\longrightarrow\sf{E=\dfrac{q}{2A\epsilon_0}}$

But electric field, $\sf{E=\dfrac{F}{q}.}$ Then,

$\longrightarrow\sf{\dfrac{F}{q}=\dfrac{q}{2A\epsilon_0}}$

$\longrightarrow\sf{F=\dfrac{q^2}{2A\epsilon_0}}$

$\longrightarrow\sf{F=\dfrac{1}{2\epsilon_0}\cdot\dfrac{q^2}{A}\quad\quad\dots(3)}$

We have, surface charge density,

$\longrightarrow\sf{\sigma=\dfrac{q}{A}}$

$\longrightarrow\sf{\sigma^2=\dfrac{q^2}{A^2}}$

$\longrightarrow\sf{\dfrac{q^2}{A}=\sigma^2A}$

$\longrightarrow\sf{\dfrac{q^2}{A}=\pi\sigma^2R^2\quad\quad\dots(4)}$

Substituting (4) in (3), we get,

$\longrightarrow\underline{\underline{\sf{F=\dfrac{\pi\sigma^2R^2}{2\epsilon_0}}}}$