Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

A capacitor is a system of two conductors separated by an insulating medium. Capacitance of this systems is determined purely geometrically, by the shapes, size and relative position of the conductors. The medium between the two conductor can be air or a dielectric. The way the dielectric is inserted accordingly by capacitance changes. Electric energy is stored in the capacitors. The capacitor can be connected in series or in parallel in an electric circuit.

The plates of a parallel plate capacitor with area of each plater A and change Q attract each other with a force.

Options:

$\frac{Q^2}{A \varepsilon_0}$

$\frac{Q^2}{2 A \varepsilon_0}$

$\frac{2 Q^2}{\varepsilon_0 A}$

$\frac{Q^2}{4 \varepsilon_0 A}$

Correct Answer:

$\frac{Q^2}{2 A \varepsilon_0}$

Explanation:

The correct answer is Option (2) → $\frac{Q^2}{2 A \varepsilon_0}$

Force between plates of a parallel plate capacitor is

F = QE

$F = Q\left(\frac{\sigma}{2 \varepsilon_0}\right) ~~~~ {\left[∵ E=\frac{\sigma}{2 \varepsilon_0}\right]}$

$F=\frac{Q^2}{2 A \varepsilon_0} ~~~~{\left[\sigma=\frac{Q}{A}\right]}$