Parallel-plate capacitor and work

[SimpliciusH]

Disclaimer: Not very good at English terminology and notation so I've improvised and googled a bit, hope this isn't *too* poorly written to understand. Please correct me on my mistakes. :)

Homework Statement

The plates of of a paralell-plate capacitor have an area of 100 cm2 and are 3 cm apart. How much work is done if we slowly part them to a distance of 5 cm, if the plates are kept at a voltage 1000V? (a) What if the plates have been isolated during the experiment? (b)

Homework Equations

capacity=influence constant*area/distance
charge= capacity * voltage
F = charge * E

The Attempt at a Solution

 

[jbsteele]

a) If the plates have been isolated during the experiment, then the work done would be equal to the change in potential energy. The potential energy of the capacitor is given by U = 1/2*CV^2, where C is the capacitance and V is the voltage applied. The capacitance of the capacitor is given by C = εA/d, where ε is the permittivity of free space, A is the area of the plates and d is the distance between them. Therefore, the change in potential energy is U = 1/2*(εA/d₁ - εA/d₂)*V², where d₁ and d₂ are the initial and final distances between the plates, respectively. For the given scenario, U = 1/2*(8.85x10⁻¹²*100/3 - 8.85x10⁻¹²*100/5)*1000² = 3.54x10⁴ J. b) If the plates are not isolated, then the work done would also include the energy dissipated in the external circuit, which is given by P = IV = I²R = V²/R, where I is the current, V is the voltage and R is the resistance of the external circuit. Therefore, the total work done is equal to the sum of the change in potential energy and the energy dissipated in the external circuit, i.e. W = U + P = 3.54x10⁴ + V²/R.