A capacitor is connected up to a 12.0 V battery and allowed to charge up. Once t

Question

A capacitor is connected up to a 12.0 V battery and allowed to charge up. Once the capacitor is fully charged, the battery is removed and a material with a dielectric constant is removed from the region between the two plates. From the quantities listed below, select all those items that decrease.

a) The capacitance of the capacitor

b) The potential difference between the plates

c) The charge on the plates

d) The electric potential energy stored by the capacitor

e) The electric field between the plates

f) None of the above

Explanation / Answer

a) Capacitance increases with dielectric constant [C = ke0A/d]
b) the capacitance: C = Q/V , the potential difference V decreases.
c) Charge is constant
d) Electric Potential Energy decreases [U = (1/2)Q2/C]

e) Electric field decreases because the potential difference decreases with dielectric. (E = V/d)

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