Question

An oil drop ‘B’ has charge 1.6 × 10−19 C and mass 1.6 × 10−14 kg. If the drop is in equilibrium position, let 10 k be the potential diff. between the plates. [The distance between the plates is 100 mm] Then what is the value of k.

Answer 1

An oil drop ‘B’ has charge $1.6 \times 10^{-19} \mathrm{C}$ and mass $1.6 \times 10^{-14} \mathrm{kg}$. If the drop is in equilibrium position, let $10 \mathrm{k}$ be the potential diff. between the plates. [The distance between the plates is 100 mm] Then what is the value of k.

Solution

In this case we can see the equilibrium between two forces: weight and electrostatic force.

$$m g = F _ {e l} = q E,$$

where $m$ – mass of drop, $g$ – gravity constant, $q$ – charge of drop.

For a flat capacitor:

$$E = \frac {V}{d},$$

where $V$ – the potential difference between the plates, $d$ – the distance between the plates.

Now we have:

$$m g = q \frac {V}{d} = 1 0 * \frac {q * k}{d}.$$

So for constant k:

$$k = \frac {1}{1 0} \frac {m * g * d}{q} = \frac {1}{1 0} \frac {1 . 6 \times 1 0 ^ {- 1 4} * 9 . 8 * 1 0 0 * 1 0 ^ {- 3}}{1 . 6 \times 1 0 ^ {- 1 9}} = 9. 8 * 1 0 ^ {3} V = 9. 8 k V.$$

Answer: 9.8 kV.