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.

A dielectric slab and a conducting slab of thickness d is inserted inside the charged capacitor plates. k is the dielectric constant. The electric field in the regions 2, 3, 4 are.

Options:

$\frac{\sigma}{\varepsilon_0}, \frac{\sigma}{k \varepsilon_0}$

$\frac{\sigma}{\varepsilon_0}, \frac{\sigma}{\varepsilon_0}, \frac{\sigma}{k \varepsilon_0}$

zero, $\frac{\sigma}{\varepsilon_0}, \frac{\sigma}{k \varepsilon_0}$

zero, zero, $\frac{\sigma}{k \varepsilon_0}$

Correct Answer:

zero, $\frac{\sigma}{\varepsilon_0}, \frac{\sigma}{k \varepsilon_0}$

Explanation:

The correct answer is Option (3) → zero, $\frac{\sigma}{\varepsilon_0}, \frac{\sigma}{k \varepsilon_0}$

We known

Electric field inside a conductor is zero.

And, if electric field in free space is E0

Then electric field in dielectric (having dielectric constant K) = $\frac{E_0}{K}$

∴   For region (2) →  $E_2$ = Zero

For region (3) →  $E_3=\frac{\sigma}{\varepsilon_0}$

For region (4) →  $E_4=\frac{\sigma}{K \varepsilon_0}$