Question

The particle P shown in figure (31-E11) has a mass of 10 mg and a charge of −0⋅01 µC. Each plate has a surface area 100 cm² on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?

Answer 1

The particle is balanced when the electrical force on it is balanced by its weight.
Thus,
$$\begin{gathered} mg=qE \\ mg=q\times \frac{V\text{'}}{d}...\left(\mathrm{i}\right) \end{gathered}$$
Here,
d = Separation between the plates of the capacitor
V' = Potential difference across the capacitor containing the particle

We know that the capacitance of a capacitor is given by
$$\begin{gathered} C=\frac{{\in }_{0}A}{d} \\ \Rightarrow d=\frac{{\in }_{0}A}{C} \end{gathered}$$
Thus, eq. (i) becomes
$$\begin{gathered} mg=q\times V\text{'}\times \frac{\mathrm{C}}{{\in }_{0}A} \\ \Rightarrow V\text{'}=\frac{mg{\in }_{0}A}{q\times C} \\ \Rightarrow V\text{'}=\frac{{10}^{-6}\times 9.8\times (8.85\times {10}^{-12})\times (100\times {10}^{-4})}{(0.01\times {10}^{-6})\times (0.04\times {10}^{-6})} \\ \Rightarrow V\text{'}=21.68\mathrm{mV} \end{gathered}$$

Since the values of both the capacitors are the same,
V = 2V' = 2 $\times$ 21.86 $\approx$ 43 mV