Three capacitors are connected as shown in the diagram. A 100.0 volt potential d

Question

Three capacitors are connected as shown in the diagram. A 100.0 volt potential difference is applied to the combination at points A and C. Find the following.

a. V1

b. V2

c. V3

d. q1

e. q2

The combination is then disassembled by removing the battery and the third capacitor (C3) at point B. This is done without disturbing the charges on the plates of the capacitors C1 and C2. Next, a dielectric (? = 5.4) is placed between the plates of capacitor C1 (the new configuration is shown below). This will rearrange the charges on the two capacitors.

f. What is the new capacitance of the system?

g. What is the potential difference across each capacitor?

i. V1

ii. V2

h. What is the charge on each capacitor?

i. q1

ii. q2

i. Numerically compare the electrical potential energy stored in capacitors 1 and 2 in the first drawing to the electrical potential energy stored in capacitors 1 and 2 in the second drawing.

ANSWERS : a)33.3 V b)33.3 V c)67 V d) 200 microC e) 133 microC f) 36.4 V

g) 1) 9V 2) 9V

h) 1) 292 microC 2) 36 microC

I) Uinitial 5.54 * 10^10^-3J, U final 1.5* 10^-3

Explanation / Answer

Here,

for the net capacitance

C1 and C2 in parallel

C4 = 6 + 4 = 10 uF

for C3 in series

1/Ceq = 1/10 + 1/5

Ceq = 3.33 uF

a) V1 = Q3/C4

V1 = 3.33 * 100/(10)

V1 = 33.3 V

b)

V2 = V1 (as both are in parallel)

V2 = 33.3 V

c) V3 = Q3/C3

V3 = 3.33 * 100/(5)

V3 = 66.7 V

d)

charge q1 = 6 * 33.3 uC

charge q1 = 200 uC

e)

charge q2 = 4 * 33.3 uC

charge q2 = 133.2 uC

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please post other in the seperate question

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