Question

Capacitor A was measured to have a capacitance of 80nF. If the charge it could hold is 30mC and the plates have a distance of 0.5cm a.) find the voltage that the capacitor was run through and b.) find thr electrice field between theplates. If the circuit diagram below only used a capacitor with the same capacitance as capaticor A. c.) find the total capacitance of the circuit

Answer 1

a) the voltage that the capacitor was run through is related to the charge it could held and its capacitance as follows:

$$U = \frac{q}{C} = \frac{30 \cdot 10^{-3}}{80 \cdot 10^{-9}} \approx 3.8 \cdot 10^5 \, V$$

b) the electric field is related to the voltage and distance between the plates as follows:

$$E = \frac{U}{d} = \frac{3.8 \cdot 10^5}{5 \cdot 10^{-3}} \approx 7.6 \cdot 10^7 \, V/m$$

c) unfortunately, the circuit diagram is not provided in the question, so I can only guess that either parallel or serial connection of two identical capacitors is implied. As a result, the total capacitance of the circuit in these cases is:

$$C_{\parallel} = C + C = 2C =160 \, nF\\ C_{ser} = \frac{C \cdot C}{C + C} = \frac{C}{2} = 40 \, nF$$

Answer: 3.8*10⁵ V, 7.6*10⁷ V/m, C_par = 160 nF, C_seq = 40 nF.