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) = -q0 * e^(-t/RC) / (RC)
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Part A = F(V) = 1/2 * Aε0(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?
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) * r2 * ε0 * 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...