Question

Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the other causes the electric field strength to be 2.00 ×105n/c. what are the diameters of the disks?

Answer 1

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

$Q=ne$

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

$Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C$

The potential difference V between the disks separated by a distance d is given by,

$V=Ed$

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

$V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V$

The capacitance C of the capacitor is given by,

$C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F$

The capacitance of a parallel plate capacitor is given by,

$C=\frac{\epsilon_0A}{d}$

Here, ε₀ is the permittivity of free space  and A is the area of the disks.

Rewrite the expression for A.

$C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3$

the area A of the disks is given by,

$A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }$

Here, D is the diameter of the disk.

$D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm$

The diameter of each disc is found to be 1.32 cm.