The capacitance of an empty capacitor is 9.30 F. The capacitor is connected to a

Question

The capacitance of an empty capacitor is 9.30 F. The capacitor is connected to a 12-V battery and charged up. With the capacitor connected to the battery, a slab of dielectric material is inserted between the plates. As a result, 8.40 × 10-5 C of additional charge flows from one plate, through the battery, and onto the other plate. What is the dielectric constant of the material?

The capacitance of an empty capacitor is 9.30 The capacitor is connected to a 12-V battery and charged up. With the capacitor connected to the battery, a slab of dielectric material is inserted between the plates. As a result, 8.40 x 10-5 C of additional charge flows from one plate, through the battery, and onto the other plate. What is the dielectric constant of the material?

Explanation / Answer

C = 9.3 uF

V = 12 V

Charge on the capacitor, Q = C*V

Now, when a dielectric slab of dielectric constant k is inserted , the Capacitance changes to k*C

So, charge now, changes to Q' = k*C*V = k*(CV)

So, change of charge = Q' - Q = CV*(k -1) = 8.4*10^-5 C

So, 9.3*10^-6*12*(k - 1) = 8.4*10^-5

So, k = 1.75 <--------- answer

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