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If a capacitor has opposite 4.9 μc charges on the plates, and an electric field of 2.4 kv/mm is desired between the plates, what must each plate's area be?

User Ssbsts
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1 Answer

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Final answer:

To find the required area of the plates of a capacitor with a charge of 4.9 μC and a desired electric field of 2.4 kV/mm, use the equation E = Q / (ε_0 * A) to get A = 0.2307 m^2.

Step-by-step explanation:

To find the area of the plates of a capacitor with a given charge and desired electric field, we can use the relationship between electric field (E), charge (Q), and area (A), as described by the equation

E = Q / (ε_0 * A),

where ε_0 is the permittivity of free space (ε_0 = 8.85 × 10^-12 C^2/N·m^2). With a charge of 4.9 μC (or 4.9 × 10^-6 C) and a desired electric field of 2.4 kV/mm (or 2.4 × 10^6 V/m), we can rearrange the equation to solve for A:

A = Q / (ε_0 * E)

Substituting the values into the equation:

A = (4.9 × 10^-6 C) / (8.85 × 10^-12 C^2/N·m^2 * 2.4 × 10^6 V/m)

A = (4.9 × 10^-6) / (2.124 × 10^-5) m^2

A = 0.2307 m^2

Therefore, the required area for each plate to achieve the desired electric field is 0.2307 m^2.

User Prabodh Tapke
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