Question

Two parallel conducting plates of equal areas A are separated by a small distance d I hey are charged to a potential difference Vby connecting a battery between them. Use these numbers: A 3.00 x 10 2 m.d-0.01cm and V- 42.0 V Charge +O Plate area .A -o a) Find the electric field between the plates. (Hint remember that the units on the electric field can be either N/C or V/m.) b) Use the value ofthe Efield to findthe surface charge density ơ on each plate. (Check your formula sheet.) Electric field E Plate separation d c) Use the value of σ to find the magnitude of the total charge O on each plate. Since there is no net charge in the circuit, then these charges must be equal in magnitude and opposite in sign. d) Take the ration of Q/V. Attach units of F for Farads to this number. Congratulations, you have found the capacitance e) If you change the voltage, does the capacitance change? Why or why not? (Try doing each step above using the formula rather than the numbers to derive a formula for the capacitance.)

Answer 1

d. O.o1x0 E 4.2x10 N = 3,72 x 10-9 C/m 2.