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,
[tex]Q=ne[/tex]
Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.
[tex]Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C[/tex]
The potential difference V between the disks separated by a distance d is given by,
[tex]V=Ed[/tex]
here, E is the electric field.
Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.
[tex]V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V[/tex]
The capacitance C of the capacitor is given by,
[tex]C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F[/tex]
The capacitance of a parallel plate capacitor is given by,
[tex]C=\frac{\epsilon_0A}{d}[/tex]
Here, ε₀ is the permittivity of free space and A is the area of the disks.
Rewrite the expression for A.
[tex]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[/tex]
the area A of the disks is given by,
[tex]A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }[/tex]
Here, D is the diameter of the disk.
[tex]D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm[/tex]
The diameter of each disc is found to be 1.32 cm.
The diameter of the disk at the given electrons is 13.1 mm.
The total charge of the capacitor is calculated as follows;
[tex]Q = nq\\\\Q = 1.5\times 10^9 \times 1.6 \times 10^{-19} \\\\Q = 2.4 \times 10^{-10} \ C[/tex]
The potential difference between the disk is calculated as follows;
[tex]V = Ed\\\\V = 2\times 10^5 \times 0.5 \times 10^{-3} \\\\V = 100 \ V[/tex]
The capacitance of the capacitor is calculated as follows;
[tex]C = \frac{Q}{V} \\\\C = \frac{2.4 \times 10^{-10}}{100} \\\\C = 2.4 \times 10^{-12} \ F[/tex]
The area of the disk is calculated as follows;
[tex]C = \frac{\varepsilon _o A}{d} \\\\A = \frac{Cd}{\varepsilon _o} \\\\A = \frac{2.4 \times 10^{-12} \times 0.5 \times 10^{-3}m }{8.85 \times 10^{-12}} \\\\A = 0.000136 \ m^2[/tex]
The diameter of the disk is calculated as follows;
[tex]A = \frac{\pi d^2}{4} \\\\d= \sqrt{\frac{4A}{\pi} } \\\\d = \sqrt{\frac{4\times 0.000136}{\pi} }\\\\d = 0.0131 \ m\\\\d = 13.1 \ mm[/tex]
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