Answer:
The energy stored in a capacitor is given by:
U = (1/2) * C * V^2
where C is the capacitance and V is the voltage across the capacitor. At the instant when the resistor is dissipating electrical energy at a rate of 20 W, the power delivered to the capacitor must be equal to the power dissipated by the resistor, since the circuit is in steady state. Therefore:
P = V^2 / R = C * dV/dt * V^2
where R is the resistance, and dV/dt is the time derivative of the voltage across the capacitor. Solving for dV/dt, we get:
dV/dt = P / (C * V^2) = 20 W / (6 μF * V^2)
Since the capacitor is initially uncharged, we have V = emf = 40 V. Integrating both sides with respect to time, we get:
ln(V) - ln(V0) = -t / RC
where V0 is the initial voltage across the capacitor, and RC is the time constant. Solving for V, we get:
V = V0 * exp(-t / RC) = 40 V * exp(-t / 30 μs)
At the instant when the resistor is dissipating electrical energy at a rate of 20 W, the energy stored in the capacitor is:
U = (1/2) * C * V^2 = (1/2) * 6 μF * (40 V * exp(-t / 30 μs))^2
Substituting t = 0, we get:
U = (1/2) * 6 μF * (40 V)^2 = 4.8 mJ
So the energy stored in the capacitor at the instant when the resistor is dissipating electrical energy at a rate of 20 W is 4.8 mJ.
Regarding problem 3, the magnetic field produced by a long, straight, cylindrical wire of radius R carrying a current uniformly distributed over its cross section is given by:
B = μ0 * I / (2πr)
where μ0 is the permeability of free space, I is the current, and r is the distance from the wire. To find the locations where the magnetic field is equal to one third of its largest value, we need to solve the equation:
B = (1/3) * Bmax
where Bmax is the maximum