Question
The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be the effect on its (i) capacity (ii) charge, (iii) potential difference (iv) electric field, (v) energy stored?
Solution
Given, that the battery remains connected and a dielectric slab is inserted in between the plates.
Capacitance of the capacitor is given by-C=
i) After the introduction of dielectric slab, capacitance becomes,
where, K is dielctric constant.
Capacitance of the capacitor is given by-C=
i) After the introduction of dielectric slab, capacitance becomes,
where, K is dielctric constant.
Therefore,
Capacitance increases as K>1 always.
ii) Charge is given by Q=CV
As C increases, Q also increases.
iii) Potential difference V is constant when, battery remains connected.
iv) Electric field is given by E=
Therefore, E reduces by a factor of 1/K.
v) Energy stored in the capacitor is given by E=.
V remains constant and C increases therby, increasing the Energy of capacitor.
Capacitance increases as K>1 always.
ii) Charge is given by Q=CV
As C increases, Q also increases.
iii) Potential difference V is constant when, battery remains connected.
iv) Electric field is given by E=
Therefore, E reduces by a factor of 1/K.
v) Energy stored in the capacitor is given by E=.
V remains constant and C increases therby, increasing the Energy of capacitor.
