The potential difference between the plates of a parallel plate capacitor is 35 V and the electric field between the plates has strength of 750 V/m. If the plate area is 4.0 × 10-2 m2, what is the capacitance of this capacitor?

7.6x10^-11

To calculate the capacitance of a parallel plate capacitor, you can use the formula:

C = ε * A / d

Where:
C is the capacitance
ε is the permittivity of free space (approximately 8.85 x 10^-12 F/m)
A is the area of the plates
d is the separation distance between the plates

Given:
Potential difference (V) = 35 V
Electric field strength (E) = 750 V/m
Area (A) = 4.0 × 10^-2 m^2

First, let's calculate the separation distance (d) using the formula:

E = V / d

Rearranging the formula:

d = V / E

Substituting the given values:

d = 35 V / 750 V/m
d ≈ 0.0467 m

Now, we can calculate the capacitance (C) using the formula:

C = ε * A / d

Substituting the given values:

C = (8.85 x 10^-12 F/m) * (4.0 × 10^-2 m^2) / (0.0467 m)

Calculating:

C ≈ 3.73 x 10^-10 F

Therefore, the capacitance of the capacitor is approximately 3.73 x 10^-10 F.

To determine the capacitance of a capacitor, we can use the formula:

C = ε0 * A / d,

where C is the capacitance, ε0 is the permittivity of free space (8.85 × 10^-12 F/m), A is the area of one of the plates, and d is the distance between the plates.

In this case, we have the potential difference (V) and the electric field strength (E), but we need to find the distance between the plates (d).

We know that the electric field E is related to the potential difference V by the equation:

E = V / d.

Rearranging this equation, we find that:

d = V / E.

Substituting the values given:

d = 35 V / 750 V/m = 0.0467 m.

Now we can calculate the capacitance C:

C = ε0 * A / d = (8.85 × 10^-12 F/m) * (4.0 × 10^-2 m^2) / 0.0467 m = 7.6 × 10^-9 F.