Two parallel plate capacitors?
Question 10
Two parallel-plate capacitors, each having a capacitance of C1 = C2 = 7
μF, are connected in parallel across a 2-V battery. The parallel combination i s then disconnected from the battery and a dielectric slab of constant K = 1.8 is inserted between the plates of the capacitor C2, completely filling the gap. After the dielectric is inserted, find the potential difference across each capacitor. Your answer must contain 2 figures after the decimal point.
Correct Answer
1.43
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Answers & Comments
Q = CV and both caps have 2V so Q1 = Q2 = 14µC
Disconnect from battery and insert dielectric so that C2 = 1.8*7 = 12.6µF
The caps are in parallel so their voltages are equal:
V1=V2 so Q1/7µF = Q2/12.6µF. Cross multiply:
12.6Q1 = 7Q2 but Q1+Q2=28µC so Q2 = 28µC-Q1
Therefore:
12.6Q1 = 7(28µC - Q1) -----> Q1(12.6+7) = 19.6Q1 = 7*28 = 196
Q1 = 196/19.6 = 10µC and Q2 = 28 - 10 = 18µC
V1 = Q1/C1 = 10µC/7µF = 1.43V
V2 = Q2/C2 = 18µC/12.6µF = 1.43V
Whoever wrote this question need a refresher course on significant figures. Your original capacitance is given with 1 sig fig. Your voltage has 1 sig fig. K has 2 sig figs. I can see an answer with 2 sig figs, but certainly not three.
Charge Q, capacitance C and potential difference V are related by:
Q = CV
Effective capacitance of two 7μF capacitors in parallel =
(7 + 7) = 14 μF
Charge stored on parallel combination = (14μF * 2V) = 28 μC
Inserting dielectric will increase capacitance of C2 from 7μF to (7 * 1.8) = 12.6μF
Effective capacitance of parallel combination will therefore now be (7 + 12.6) = 19.6μF
Again, Q = CV, so:
V = Q/C = (28μC / 19.6μF) = 1.43 V to three sig figs