How strong is the electric field between the plates of a [tex]0.60 \, \mu F[/tex] air-gap capacitor if they are [tex]3.0 \, \text{mm}[/tex] apart and each has a charge of [tex]76 \, \mu C[/tex]?

Question

31-07-2024
Physics

High School

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How strong is the electric field between the plates of a [tex]0.60 \, \mu F[/tex] air-gap capacitor if they are [tex]3.0 \, \text{mm}[/tex] apart and each has a charge of [tex]76 \, \mu C[/tex]?

Answer :

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diyapatel34 diyapatel34 · Answer 1 · 2023-09-23T18:17:22-04:00

Final answer:

The electric field strength between the plates of the 0.60 μF air-gap capacitor, each with a charge of 76 μC and separated by 3.0 mm, is 4.41 x 10^5 V/m.

Explanation:

The strength of the electric field between the plates of a capacitor can be calculated using the formula E = Q / (ε₀ * A), where E is the electric field strength, Q is the charge on the plates, ε₀ is the permittivity of free space (8.85 x 10-12 C2/N·m2), and A is the area of the plates. The area A can be derived from the capacitance (C) and plate separation (d) using the formula C = ε₀ * A / d. In this case, we need to find the area using the given capacitance and separation, and then use the charge and area to find the electric field strength.

First, we calculate the area (A) as follows: A = C * d / ε₀. Plugging in the values: A = (0.60 x 10-6 F) * (3.0 x 10-3 m) / (8.85 x 10-12 C2/N·m2) = 0.2034 m2. Next, we use the area and charge (Q) to calculate the electric field (E): E = Q / (ε₀ * A) = (76 x 10-6 C) / (8.85 x 10-12 C2/N·m2 * 0.2034 m2) = 4.41 x 105 V/m.

Therefore, the electric field strength between the plates of the 0.60 μF air-gap capacitor, with each plate having a charge of 76 μC and separated by 3.0 mm, is 4.41 x 105 V/m.