Question

Two circular disks spaced 0.005 m (0.50 cm) apart form a parallel-plate capacitor. Transferring 1.10 ×10^9 electrons from one disk to the other causes the electric field strength to be 2.50 ×10^5N/C. What are the diameters of the disks? Express your answer with the appropriate units.

Answer 1

An expression for amount of charge stored on each plate of a capacitor is given by -

Q = n e

Q = [(1.10 x 10⁹ electrons) (1.6 x 10^-19 C/electrons)]

Q = 1.76 x 10^-10 C

In parallel-plate capacitor, we have

E = / ₀   = E ₀

where, E = magnitude of the electric field strength = 2.50 x 10⁵ N/C

₀ = permittivity of free space = 8.85 x 10^-12 C²/Nm²

then, we get

= [(2.50 x 10⁵ N/C) (8.85 x 10^-12 C²/Nm²)]

= 2.2125 x 10^-6 C/m²

We know that,   Q = A = (r²)

r² = Q /

r = (1.76 x 10^-10 C) / [(2.2125 x 10^-6 C/m²) (3.14)]

r = 0.0000253 m²

r = 5.03 x 10^-3 m

Therefore, diameters of the disks which will be given by -

D = 2 r

D = [2 (5.03 x 10^-3 m)]

D = 1.006 x 10^-2 m

convert m into cm :

D = 1.006 cm