The capacitance between the adjacent plates shown in figure (31-E20) is 50 nF. A charge of 1⋅0 µC is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate? (b) Find the potential difference developed between the upper and the middle plates.

Open in App

Solution

(a) When the charge of 1 µC is introduced on plate B, the charge gets equally distributed on its surface. All sides of the plate gets 0.5 µC of charge. Due to induction, 0.5 µC charge is induced on the upper surface of plate A.

(b) Given:
Capacitance, C = 50 nF = 50 × 10^{$-$ 9} F = 5 × 10^{$-$ 8} F
Effective charge on the capacitor, $Q_{net} = \frac{1 μC - 0}{2} = 0.5 μC$
∴ Potential difference between plates B and A, $V = \frac{Q_{net}}{C} = \frac{5 \times 10^{- 7}}{5 \times 10^{- 8}} = 10 V$

Suggest Corrections

Similar questions

Q. The capacitance between adjacent plates shown in Fig 20.95(a) is 50 nF.

A 1 μ C

charge is placed on the middle plate. Find the charge on the outer surface of upper plate and the potential difference between upper and middle plates:

Q. Consider the situation of the previous problem. If 1⋅0 µC is placed on the upper plate instead of the middle, what will be the potential difference between (a) the upper and the middle plates and (b) the middle and the lower plates?

A charge of $20 μ C$ is placed on the positive plate of an isolated parallel-plate capacitor of capacitance $10 μ F .$ Calculate the potential difference developed between the plates.

A charge of $+ 2.0 \times 10^{- 8} C$ is placed on the positive plate and a charge of $- 1.0 \times 10^{- 8} C$ on the negative plate of a parallel-plate capacitor of capacitance $1.2 \times 10^{- 3} μ F .$

Calculate the potential difference developed between the plates.

Join BYJU'S Learning Program