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Question: Find the force between the plates of a capacitor of charge 5mC and area of the plates 1 cm²....

Find the force between the plates of a capacitor of charge 5mC and area of the plates 1 cm².

A

14 μΝ

B

14 kN

C

14 GN

D

1.4 GN

Answer

14 GN

Explanation

Solution

The force between the plates of a parallel plate capacitor with charge QQ and area AA is given by the formula:

F=Q22Aϵ0F = \frac{Q^2}{2A\epsilon_0}

where:

  • QQ is the magnitude of the charge on each plate

  • AA is the area of each plate

  • ϵ0\epsilon_0 is the permittivity of free space. ϵ0≈8.854×10−12 C2/N⋅m2\epsilon_0 \approx 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2

Given values:

  • Charge Q=5 mC=5×10−3 CQ = 5 \, \text{mC} = 5 \times 10^{-3} \, \text{C}

  • Area A=1 cm2=1×(10−2 m)2=1×10−4 m2A = 1 \, \text{cm}^2 = 1 \times (10^{-2} \, \text{m})^2 = 1 \times 10^{-4} \, \text{m}^2

Substitute the values into the formula:

F=(5×10−3 C)22×(1×10−4 m2)×(8.854×10−12 C2/N⋅m2)F = \frac{(5 \times 10^{-3} \, \text{C})^2}{2 \times (1 \times 10^{-4} \, \text{m}^2) \times (8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2)}

F=25×10−6 C22×10−4 m2×8.854×10−12 C2/N⋅m2F = \frac{25 \times 10^{-6} \, \text{C}^2}{2 \times 10^{-4} \, \text{m}^2 \times 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2}

F=25×10−617.708×10−16 NF = \frac{25 \times 10^{-6}}{17.708 \times 10^{-16}} \, \text{N}

F=2517.708×10−6−(−16) NF = \frac{25}{17.708} \times 10^{-6 - (-16)} \, \text{N}

F=2517.708×1010 NF = \frac{25}{17.708} \times 10^{10} \, \text{N}

F≈1.4117×1010 NF \approx 1.4117 \times 10^{10} \, \text{N}

To compare with the options, let's express the result in terms of GigaNewtons (GN).

1 GN=109 N1 \, \text{GN} = 10^9 \, \text{N}

F≈1.4117×1010 N=1.4117×10×109 N=14.117×109 N=14.117 GNF \approx 1.4117 \times 10^{10} \, \text{N} = 1.4117 \times 10 \times 10^9 \, \text{N} = 14.117 \times 10^9 \, \text{N} = 14.117 \, \text{GN}

The calculated value 14.117 GN14.117 \, \text{GN} is closest to option C, 14 GN.