Chapter 9 : Capacitors and Dielectrics Example 9.4 (Level III): Two capacitors a

Question

Chapter 9: Capacitors and Dielectrics

Example 9.4 (Level III): Two capacitors are connected to each other and the battery is removed. The voltage across each capacitor is 6 volts. V = V1 = V2 = 6 V. The capacitance of C1 = 1.0 x 10 ^-6 F and the capacitance of C2= 2.0 x 10 ^-6 F. What is the charge on each capacitor? If the gap between the plates of C1 if filled with paraffin (dielectric constant K=2.2) what is the new charge on each capacitor and what is the new voltage across each capacitor? [The total charge is conserved and the voltage across the first capacitor is still equal to that across the second capacitor.] Calculate the total stored potential energy of the charged capacitors before and after the paraffin was added. Was there a loss or gain of potential energy when the dielectric was inserted? What % of the original potential energy was gained or lost?

Kean University - Dr. Anwamgh

Explanation / Answer

Given v1 = v2 =6vAnd c1 = 1* 10 power of -6 and c2 = 2 * 10 power of -6

so q1=6 x 10-6 and q2 = 12 x 10-6 by formul q = c x v

given c1 filled with dielectric so, new C1 = K x c here kk= dielectric constant

C1 = 2.2 x 10-6

Now using charge is conserved and voltage constant across sources

SO,q1 / c1 = 18 x 10-6 - q1 /c2 here q2 = 18 x 10 -6 - q1

Solving the above we get q1 = 9.4 x 10-6 and q2 = 8.6 x 10-6

We know that when fformula pot energy = 1/2 C x v

So new C1 = 2.2 x c

So potential energy increases by 220%

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