1. A parallel capacitor is connected to a constant potential (voltage) source. (

Question

1. A parallel capacitor is connected to a constant potential (voltage) source. (a) While it is connected to the source, a dielectric is introduced to fill the space between the plates. What happens to (i) the electric field between the two plates, (ii) the magnitude of the charge on each of the plates, (iii) the voltage between the two plates? (b) Now consider a situation where the capacitor is charged and then disconnected from the voltage source. A dielectric is then introduced to fill the space between the plates. What happens to (i) the electric field betweern the two plates, (ii) the magnitude of the charge on each of the plates, (iii) the voltage between the two plates?

Explanation / Answer

a)

i) electric field is given as : E = V /d since V and "d" both remains same hence electric field remains same on introduction of dielectric

ii)

charge is given as : Q = CV. as dielectric is introduced , the Capacitance "C" increases, and V remains same. hence the charge also increase since Q directly depends on "C"

iii)

V remains same since the capacitor is connected to a constant source Voltage

b)

i)

electric field decreases due to polarization in the dielectric introduced

ii)

magnitude of charge remains constant since the capacitor can not aquire more charge after being disconnected from power supply

iii)

Voltage decreases

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