Part A = Zero

Part B = Ε

Part C = clockwise

Part D = the top plate

Part E = zero

Part F = q = CΕ

Part G = W = CΕ^2

Part H = q(t) = CΕ * (1 – e^(-t/RC))

Part I = I(t) = (Ε / R) * e^(-t/RC)

Part J = I(t) = (Ε / R) * e^(-t/RC)

Part K = I(t) = -q_{0} * e^(-t/RC) / (RC)

Click for more

Click for More...

Part A = F(V) = 1/2 * Aε_{0}(V^2 / d^2)

Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.

Click for More...

Part A = Option 3

A positive charge is brought close to a fixed neutral conductor that has a cavity. The cavity is neutral; that is, there is no net charge inside the cavity.

Which of the figures best represents the charge distribution on the inner and outer walls of the conductor?

Part A = 12.1 ms

Part B = R = 1.12 kΩ

Part C = t = 10.1 ms

Part D = 595 J

A 10.8 µF capacitor in a heart defibrillator unit is charged fully by a 10500 V power supply.

Find the time constant.

Determine the resistance, R.

How much time does it take for the capacitor to lose 81 % of its stored energy?

If the paddles are left in place for many time constants, how much energy is delivered to the chest/heart area of the patient?

Part A = Individual charges flow through the circuit from the negative to the positive plate of the capacitor., Current flows clockwise through the circuit.

Part B = IP = IB = IN

Part C = decreases.

Part D = 6 s

Part A = I = ((K-1) * r^{2} * ε_{0} * V) / (d * Δt)

Assuming that the dielectric is inserted at a constant rate, find the current I as the slab is inserted.

Click for More...