Question

A capacitor is to be constructed to have a capacitance of 100µF. The area of the plates is 6.0m x 0.030m, and the relative permittivity of the dielectric is 7.0. Find the necessary separation of the plates and the electric field strength if a potential difference of 12V is applied across the capacitor.

A. Separation: 0.0024m, Electric Field Strength: 500V/m

B. Separation: 0.0024m, Electric Field Strength: 1000V/m

C. Separation: 0.0012m, Electric Field Strength: 500V/m

D. Separation: 0.0012m, Electric Field Strength: 1000V/m

Answer 1

To find the separation between the plates of a capacitor with a capacitance of 100µF and the electric field strength, we can use the equations for capacitance and electric field strength.

Step-by-step explanation:

To find the separation between the plates of a capacitor with a capacitance of 100µF, we can use the equation:

C = (ε₀εᵣA) / d

Where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10-12 F/m), εᵣ is the relative permittivity of the dielectric (7.0 in this case), A is the area of the plates (6.0m x 0.030m), and d is the separation between the plates.

Substituting the given values into the equation, we have:

100µF = (8.85 x 10-12 F/m * 7.0 * 0.010m²) / d

Simplifying, we can solve for d:

d = (8.85 x 10-12 F/m * 7.0 * 0.010m²) / 100µF = 0.0024m

The electric field strength can be calculated using the equation:

E = V / d

Where E is the electric field strength, V is the potential difference (12V in this case), and d is the separation between the plates.

Substituting the given values into the equation, we have:

E = 12V / 0.0024m = 500V/m

Therefore, the necessary separation of the plates is 0.0024m and the electric field strength is 500V/m.