Two metal plates **(A, B)** are kept horizontally with separation of $\left(\dfrac{12}{\pi}\right)$ **cm**, with plate A on the top.
An atomizer jet sprays oil (density **1.5 g/cm³**) droplets of radius **1 mm** horizontally. All oil droplets carry a charge **5 nC**. The potentials $V_A$ and $V_B$ are required on plates A and B respectively in order to ensure the droplets do not descend. The values of $V_A$ and $V_B$ are ______.
(Neglect the air resistance to the droplets and take **g = 10 m/s²**)
Answer: (A) $100$ V and $580$ V
Mass of a droplet: $m=\rho\cdot\tfrac43\pi r^3=1500\times\tfrac43\pi\times10^{-9}=2\pi\times10^{-6}$ kg.
For the droplet not to descend, $qE=mg$:
$$E=\frac{mg}{q}=\frac{2\pi\times10^{-6}\times10}{5\times10^{-9}}=4\pi\times10^{3}\ \text{V/m}$$
Potential difference between the plates: $V=Ed=4\pi\times10^{3}\times\dfrac{0.12}{\pi}=480$ V.
The charge is positive, so the field must point upward, from B (bottom) to A (top). Hence $B$ must be at the higher potential: $V_B-V_A=480$ V.
Only $V_A=100$ V, $V_B=580$ V satisfies this.
Solution by Sreeraj P, M.Sc Physics