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

Answer 1

The correct option is A

$v^2=2y\left[\frac{qQ}{4\pi {\epsilon }_{0}R\left(R+y\right)m}+g\right]$

Step 1 : Given data

Radius of the sphere $=R$

Mass of the piece $=m$

Charge of the piece $=q$

Speed $=v$

Height $=y$

Step 2 : To find the speed

By using law of conservation of energy

$$\begin{gathered} ∆KE+{\left(∆PE\right)}_{Electrical}+{\left(∆PE\right)}_{gravitational}=0 \\ \frac{1}{2}m{v}^2+\left(k\frac{Qq}{R+y}-k\frac{Qq}{R}\right)+\left(-mgy\right)=0 \\ \frac{1}{2}m{v}^2=mgy+kQq\left(\frac{1}{R}-\frac{1}{R+y}\right) \\ {v}^2=2gy+\frac{2kQq}{m}\frac{y}{R\left(R+y\right)} \\ {v}^2=2y\left[\frac{qQ}{4\pi {\epsilon }_{0}R\left(R+y\right)m}+g\right] \end{gathered}$$

Hence the speed $v^2=2y\left[\frac{qQ}{4\pi {\epsilon }_{0}R\left(R+y\right)m}+g\right]$

Therefore Option A is the correct answer