Question

The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the
(A) Decreases
(B) Remains unchanged
(C) Becomes zero
(D) Increase

Answer 1

Hint: We know that the voltage difference between any two points in a circuit is known as Potential Difference and it is this potential difference which makes current flow. Unlike current which flows around a closed electrical circuit in the form of electrical charge, potential difference does not move or flow when it is applied. Electrical potential difference is the difference in the amount of potential energy a particle has due to its position between two locations in an electric field. This important concept provides the basis for understanding electric circuits. Based on this concept we have to solve this question.

Complete step-by-step answer:
At first, we know that capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. There are two closely related notions of capacitance: self-capacitance and mutual capacitance. Any object that can be electrically charged exhibits self-capacitance.
In general capacitance of parallel plate capacitor is given by:
$C=\dfrac{k{{\epsilon }_{0}}A}{d}$
Where C is capacitance, k is relative permittivity of dielectric material, ${{\epsilon }_{0}}$
is permittivity of free space constant. A is the area of plates and d is the distance between them.
We should know that relative permittivity is the factor by which the electric field between the charges is decreased relative to vacuum. Likewise, relative permittivity is the ratio of the capacitance of a capacitor using that material as a dielectric, compared with a similar capacitor that has vacuum as its dielectric.
Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material. Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.

Hence, the correct answer is Option A.

Note: We know that the electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field. When we talked about electric fields, we chose a location and then asked what the electric force would do to an imaginary positively charged particle if we put one there.
We know that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if we think of the change in the potential energy as we bring the two charges closer or move them farther apart.