A parallel-plate capacitor with a capacitance of 1.00 x102 F is initially unchar
ID: 1499642 • Letter: A
Question
A parallel-plate capacitor with a capacitance of 1.00 x102 F is initially uncharged. It is then connected to 1.00 x102 V battery and a switch in the circuit is then closed.
(A) Draw the circuit, including the direction the current will flow until the capacitor is charged.
(B) How much charge arrives on each plate of the capacitor? Indicate these values on your sketch.
(C) What is the potential difference across the plates of the capacitor?
(D) What is the magnitude and direction of the E-Field between the plates of the capacitor?
The battery is then removed, and the circuit is closed. A sheet of teflon (dielectric constant = 2.00) is then slid between the plates of the capacitor.
(E) What is the new charge on the plates?
(F) What is the new magnitude of the E-Field between the plates?
(G) What is new potential difference between the plates?
Explanation / Answer
A) Direction of the current is from the positive terminal of battery to the capacitor and than returning to the negative terminal of battery .
B) charge arrives on each plate of the capacitor = CV
= (1.00 x102 * 10-6) * 1.00 x102
= 0.01 C
C) potential difference across the plates of the capacitor = 100 V
D) magnitude of the E-Field between the plates = 100/10-3
= 105 V/m
Direction of Electric field from positive plate to negative plate .
E) new charge on the plates = 2.00 * 0.01 = 0.02 C
F) new magnitude of the E-Field between the plates = 105 V/m
G) new potential difference between the plates = 100 V
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