Question 412: Electrostatics & Capacitance

Question

A solid sphere of radius R carries a charge
Q + q distributed uniformly over its volume. A
very small point like piece of it of mass m gets
detached from the bottom of the sphere and
falls down vertically under gravity. This piece
carries charge q. If it acquires a speed v when
it has fallen through a vertical height y (see
figure), then :
(assume the remaining portion to
be spherical).
![](https://imagex.cdn.examgoal.net/jgy2xukfg6ux0d/7b8f4f3d-bc30-4aa6-ba14-b114e09b5257/6ff0b4d0-fe0c-11ea-8293-3b6cc591631c/file-jgy2xukfg6ux0e-272w.png)

Accepted Answer

Applying mechanical energy conservation,
${K_A} + {U_A} = {K_B} + {U_B}$
$ \Rightarrow $ $0 + mgy + q{V_A} = {1 \over 2}m{v^2} + 0 + ( + q{V_B})$
$ \Rightarrow $ $mgy + q{V_A} = {1 \over 2}m{v^2} + q({V_B})$
$ \Rightarrow $ $mgy + {{qk(Q)} \over R} = {1 \over 2}m{v^2} + {{qk(Q)} \over {R + y}}$
$ \Rightarrow $ ${1 \over 2}m{v^2} =  - {{kq(Q)} \over {R + y}} + {{kq(Q)} \over R} + mgy$
$ \Rightarrow $ ${{m{v^2}} \over 2} = {{ - kq(Q)R + kq(Q)(R + y)} \over {R(R + y)}} + mgy$
$ \Rightarrow $ ${v^2} = {2 \over m}\left[ {{{ - kqQR + kqQR + kqQ} \over {R(R + y)}} + mgy} \right]$
$ \Rightarrow $ ${v^2} = {2 \over m}\left[ {{{kq(Q)y} \over {R(R + y)}} + mgy} \right]$
$ \Rightarrow $ ${v^2} = 2y\left[ {{{q(Q)} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right] = 2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right]$
![](https://imagex.cdn.examgoal.net/1kjtowt6z/1799b007-f5fc-48f8-aca0-dcd6f7a49a33/16e4a9b0-54aa-11eb-bc03-4b67b2efa0ee/file-1kjtowt70-376w.png)

Answer

v
2
= $y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R\left( {R + y} \right)m}} + g} \right]$

Answer

v
2
= $2y\left[ {{{qQR} \over {4\pi {\varepsilon _0}{{\left( {R + y} \right)}^3}m}} + g} \right]$

Answer

v
2
= $2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R\left( {R + y} \right)m}} + g} \right]$

Answer

v
2
= $y\left[ {{{qQ} \over {4\pi {\varepsilon _0}{R^2}ym}} + g} \right]$

Electrostatics & Capacitance

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