In this example, we analyze a system of two parallel plates, each charged with 30 nano coulombs (30 x 10^-9 coulombs), separated by a distance of 10 millimeters (0.01 meters). The first task is to determine the voltage (potential difference) between the plates. The relationship between charge (q), capacitance (C), and voltage (V) is given by the formula:

q = C * V

From this, we can express voltage as:

V = q / C

To find the capacitance of the parallel plate capacitor, we use the formula:

C = ε₀ * (A / d)

Here, ε₀ is the permittivity of free space, approximately 8.85 x 10^-12 F/m, A is the area of the plates, and d is the separation distance. The area of the plates, each measuring 1 cm x 1 cm, is converted to square meters:

A = 0.01 m * 0.01 m = 1 x 10^-4 m²

Substituting the values into the capacitance formula gives:

C = 8.85 x 10^-12 * (1 x 10^-4 / 0.01) = 8.85 x 10^-12 * 1 x 10^-2 = 8.85 x 10^-14 F

Now, substituting the charge and capacitance back into the voltage formula yields:

V = (30 x 10^-9) / (8.85 x 10^-14) ≈ 3.39 x 10⁵ volts

Next, we calculate the electric field (E) between the plates, which is defined as the voltage divided by the distance:

E = V / d

Substituting the known values gives:

E = (3.39 x 10⁵) / (0.01) = 3.39 x 10⁷ N/C

Finally, we determine the energy required to move a negative charge of -5 nano coulombs (5 x 10^-9 coulombs) from the positive plate to the negative plate. The work done (W) in moving a charge across a potential difference is given by:

W = -q * ΔV

In this case, ΔV is the potential difference, and since we are moving a negative charge from a higher to a lower potential, we have:

W = -(-5 x 10^-9) * (3.39 x 10⁵) = 1.695 x 10^-3 joules

The negative sign indicates that work is done on the system, meaning energy must be supplied to move the charge against the electric field. Thus, the energy required to move the charge is approximately 1.7 x 10^-3 joules.