Question

A dielectric is pulled out from between the plates of a capacitor which remains connected to a battery. What changes occur to (a) the capacitance, (b) the charge on the plates, (c) the potential difference, (d) the energy stored in the capacitor, and (e) the electric field? Explain your answers.

Short answer

(a) The capacitance of capacitor decreases.

(b) The charge on plates decreases.

(c) The potential difference stays the same.

(d) The energy stored reduces.

(e)The electric field remains constant.

Step-by-step solution

Understanding dielectric constant

The dielectric constant is described as the term which is equivalent to the fraction of permittivity of a material and permittivity of a vacuum.

(a) Evaluation of the change in capacitance

The relation of capacitance is given by,

\(C = \frac{{K{\varepsilon _0}A}}{d}\)

Here, Ais the area, Cis the capacitance,\({\varepsilon _0}\)is the permittivity of vacuum, Kis the dietetic constant and dis the distance between the plates.

The value of dielectric constant is always\(K > 1\).

In the above relation, it is seen that as the dielectric is removed, the value of capacitance reduces by a factor of K.

(b) Evaluation of the change in charge of the plates

The relation of charge is given by,

\(Q = CV\)

Here, Cis the capacitance, Qis the charge and Vis the voltage.

The charge on plates relies on the capacitance and voltage across the capacitance. As the capacitance reduces, the charge will also decrease.

(c) Evaluation of the change in potential difference

As given in the question, the plates of a capacitor are still connected to the same battery before and after removing the dielectric. Hence, the potential difference will remain the same.

(d) Evaluation of the change in energy stored in the capacitor

The relation of energy stored is given by,

\(E = \frac{1}{2}C{V^2}\)

Since the potential difference is still the same and capacitance reduces, the energy stored in the capacitor will reduce. It relies on capacitance, and due to a reduction in capacitance, the energy stored must decrease.

(e) Evaluation of the change in electric field

The relation of electric filed is given by,

\(V = {E_{\rm{f}}}d\)

Here,\({E_{\rm{f}}}\)is the electric field and dis the distance between the plates.

The above part shows that the potential difference remains constant, and the distance between the plates does not change. Hence, the electric field will remain the same.