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(A) Let $mg$ be the weight of the drop and $F_v = 6\pi \eta rv$ be the viscous drag force.
$1$. When the drop falls with terminal velocity $V$ in the

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$V/2$

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(A) Let $mg$ be the weight of the drop and $F_v = 6\pi \eta rv$ be the viscous drag force.
$1$. When the drop falls with terminal velocity $V$ in the absence of an electric field, the forces are balanced:
$mg = 6\pi \eta rV$ --- $(i)$
$2$. When an electric field $E$ is applied upward, the drop moves upward with terminal velocity $2V$. The electric force $QE$ acts upward, while gravity $mg$ and viscous drag $6\pi \eta r(2V)$ act downward:
$QE = mg + 6\pi \eta r(2V)$
Substituting $mg = 6\pi \eta rV$ from $(i)$:
$QE = 6\pi \eta rV + 12\pi \eta rV = 18\pi \eta rV$ --- $(ii)$
$3$. When the electric field is reduced to $E/2$, let the new terminal velocity be $V'$. Since $QE/2 = (18\pi \eta rV)/2 = 9\pi \eta rV$, the electric force is now $9\pi \eta rV$.
If the drop moves upward, the force balance is:
$QE/2 = mg + 6\pi \eta rV'$
$9\pi \eta rV = 6\pi \eta rV + 6\pi \eta rV'$
$3\pi \eta rV = 6\pi \eta rV'$
$V' = V/2$
Since $V'$ is positive, the drop moves upward with terminal velocity $V/2$.

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