A simple circuit contains only a parallel plate capacitor and a battery. The cap

Question

A simple circuit contains only a parallel plate capacitor and a battery. The capacitor is
charged fully by a 12 V battery. The parallel surface is a 900 cm^2 square. The plates are
infinitesimally thin. The gap between the plates is 0.5 cm. A dielectric, with dielectric
constant =1.5 , is then put in between the plates causing the plates to triple the distance
between them. After the distance is tripled, the battery is disconnected. [Everything else is the
same for the problem.]
Determine if the following increase, decrease, stay the same, or can not be determined by the
information given when comparing the values before the dielectric is added to after the battery
is disconnected. Explain your reasoning.
(a) Voltage between the plates
(b) Total electric field strength between the plates
(c) Magnitude of charge on the plates
(d) Electric field strength between the plates from the plates
(e) The electric potential of the negatively charged plate

Explanation / Answer

a.) the voltage remains the same between the plates. The battery is disconnected after inserting the dielectric. till then it remains connected so voltage remains unchanged.

b.) E = V/d

V is constant and d becomes three times so E decreases.

inside the dielectric Elecric field becomes 'k' times.............k is the dielectric constant

so, net field becomes k/d = 0.5 times

decreases

c.) Q = CV, V remains constant

so A is proportional to C

C is proportional to k/d, where k is the dielectric constant and d is the distance betweeen the plates.

s= C becomes (1.5/3) = 1/2 times, i.e half of previous value

so, charge decreases

d.) Elecreic field strength between the plates from the plates (outside the dielectric, k=1) becomes 1/d times = 1/3 times

hence decreases ...... explained in part (b)

e.)It remains the same as the voltge is the same.

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