Capacitor Energy and Charge Calculation

How can we calculate the total charge on each plate of a capacitor?

Given that it takes 18 J of energy to move a 0.30-mC charge from one plate of a 15-μF capacitor to the other, how can we determine the charge on each plate?

Answer:

The total energy stored in this capacitor can be used to compute the voltage across its plates. Then, using the definition: Charge = Capacitance x Voltage, the total charge is calculated. As capacitors hold equal but opposite charges, this total charge is equally distributed among the two plates.

To answer this question, you have to rely on two physics concepts: the measurements of energy and the definition of a capacitor. Energy in joules can be calculated as half the product of the capacitance (in farads) and the square of the voltage (in volts). The formula is given by E = 0.5*C*V^2. The capacitor is an electric device with two plates holding equal but opposite charges.

Given that energy E=18J and capacitance C=15*10^-6 F (from μF to F), we can substitute these into the energy formula and solve for voltage. V (volt) being the square root of (2*E/C). Thus, V=sqrt((2*18)/(15*10^-6)) =~ 7766.61 Volts.

Now that we have the voltage, we can use the definition of a capacitor which is Q=C*V to find the charge Q. In this case, Q equals 0.30mC * 7766.61 volts ~= 0.116 C. This is the total charge on the capacitor, with each plate having approximately 0.058C charge.