SATHEE: Electrostatics 5 Question 35

Electrostatics 5 Question 35

37. A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its $Q_{1}$ plates that covers $1 / 3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_{1}$ . When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ . The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option/options, ignoring edge effects.

(a) $\frac{E_{1}}{E_{2}} = 1$

(b) $\frac{E_{1}}{E_{2}} = \frac{1}{K}$

(d) $\frac{C}{C_{1}} = \frac{2 + K}{K}$

(2014 Adv.)

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Answer:

Correct Answer: 37. (a,d)

Solution:

$\begin{aligned} C & = C_{1} + C_{2} \\ C_{1} & = \frac{K ε_{0} A / 3}{d} \\ C_{2} & = \frac{ε_{0} 2 A / 3}{d} \\ \Rightarrow C & = \frac{(K + 2) ε_{0} A}{3 d} \\ \Rightarrow \frac{C}{C_{1}} & = \frac{K + 2}{K} \end{aligned}$

Also, $E_{1} = E_{2} = V / d$ , where $V$ is potential difference between the plates.

Electrostatics 5 Question 35

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Electrostatics 5 Question 35

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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