Question

In Millikan oil drop experiment a charged drop falls with a terminal velocity $V$. If an electric field $E$ is applied vertically upwards it moves with terminal velocity $2V$ in upward direction. If electric field reduces to $E/2$ then its terminal velocity will be-

• A. $\frac{V}{2}$
• B. $V$
• C. $\frac{3V}{2}$
• D. $2V$

Answer 1

The correct option is C $\frac{3V}{2}$

In absence of electric field (i.e. $\epsilon =0$)

$Mg=6\pi nv$

$D_1=6\pi nrv⟶\left(1\right)$

In presence of electric field

$Mg+QE=6\pi n\left(2V\right)⟶\left(2\right)$

When electric field $D$ reduced to $E/2$

$Mg+Q\left(E/2\right)=6\pi n\left(V^1\right)$

$D_3=6\pi nr\left(2V\right)⟶\left(3\right)$

After solving $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$

We get $V^1=\frac{3}{2}V$.