SATHEE: Electrostatics 5 Question 16

Electrostatics 5 Question 16

18. A parallel plate capacitor of capacitance $90 pF$ is connected to a battery of emf $20 V$ . If a dielectric material of dielectric constant $K = 5 / 3$ is inserted between the plates, the magnitude of the induced charge will be

(2018 Main)

(a) $0.9 μ C$

(b) $1.2 μ C$

(d) $2.4 μ C$

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Answer:

Correct Answer: 18. (b)

Solution:

$\begin{aligned} Q & = K C_{0} V \\ | Q_{in} | & = | Q 1 - \frac{1}{K} | \\ = (90 \times 10^{- 12}) (20) \frac{5}{3} 1 - \frac{3}{5} C \\ = 1200 \times 10^{- 12} C = 1.2 μ C \end{aligned}$

Electrostatics 5 Question 16

18. A parallel plate capacitor of capacitance $90 pF$ is connected to a battery of emf $20 V$ . If a dielectric material of dielectric constant $K = 5 / 3$ is inserted between the plates, the magnitude of the induced charge will be

Answer:

Solution:

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Electrostatics 5 Question 16

18. A parallel plate capacitor of capacitance 90pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K=5/3 is inserted between the plates, the magnitude of the induced charge will be

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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18. A parallel plate capacitor of capacitance $90 pF$ is connected to a battery of emf $20 V$ . If a dielectric material of dielectric constant $K = 5 / 3$ is inserted between the plates, the magnitude of the induced charge will be