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).

• A. $v^2=y[\frac{qQ}{4\pi {ϵ}_0{R}^{2}ym}+g]$
  
• B. $v^2=y[\frac{qQ}{4\pi {ϵ}_{0}R(R+y)m}+g]$
• C. $v^2=2y[\frac{QqR}{4\pi {ϵ}_0(R+y{)}^{3}m}+g]$
• D. $v^2=2y[\frac{qQ}{4\pi {ϵ}_{0}R(R+y)m}+g]$

Answer 1

The correct option is D $v^2=2y[\frac{qQ}{4\pi {ϵ}_{0}R(R+y)m}+g]$

By using energy conservation,
$\Delta KE+(\Delta PE{)}_{\text{Electro}}+(\Delta PE{)}_{\text{gravitational}}=0$
$\frac{1}{2}m{v}^2+(k\frac{Qq}{R+y}-k\frac{Qq}{R})+(-mgy)=0$
$\Rightarrow \frac{1}{2}m{v}^2=mgy+kQq(\frac{1}{R}-\frac{1}{R+y})$
$\Rightarrow v^2=2gy+\frac{2kQq}{m}\frac{y}{R(R+y)}$
$\Rightarrow v^2=2y[\frac{qQ}{4\pi {ϵ}_{0}R(R+y)m}+g]$