# 15 - Capacitors Questions Answers

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**Submit By: MANISH SIR**5 year ago

What is the main priniciple strore energy in a capacitor ?

**Asked By: DIBYA JYOTI DEKA**3 year ago

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**A capacitor of capacity 0.1 ****μF connected in series to a resistor of 10 MΩ is charged to a certain potential and then made to discharge through resistor. The time in which the potential will take to fall to half its original value is ****(Give, log _{10}2 = 0.3010)**

**Asked By: SAUMYA PALIWAL**4 year ago

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When An external resistance R is connected with the battery of emf E and internal resistane r, what is the maximum power?

**Asked By:**5 year ago

**Solved By: SARIKA**

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distance b/w plate of capacitor C &q charge is increased to double . work done=??

as distance d is double thus C= Ae/2d thus capacitance will be halved so wd shld be (1/2 )q^{2}/(C/2) ie 1/4 q^{2}/C

BUT Sir the ans given is=( 1/2)q^{2}/C Plzz do me a little favour by solving this query

**Asked By: SARIKA SHARMA**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

(1/2 )q^{2}/(C/2) = q^{2}/C not 1/4 q^{2}/C

so work done = new energy - previous energy

a II plate capacitor has an electric field of 10^{5} V/m b/w d plates . if charge on capacitor plate is 1uC d force on each capacitor is??

ans 0.05N

**Asked By: SARIKA SHARMA**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

F = σq/2ε_{0} = Eq/2 = 10^{5}*10^{-6}/2 = 0.05

A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates. Please add explanation!

**Asked By: VISHWAJEET MISHRA**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

due to induction the charges on the four faces will be

so V = q/C = 1.5 * 10^{-8}/1.2 * 10^{-9}

solve now

**Asked By: GAURAV MAHATE**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

according to the diagram

tanθ = qE/mg = q(σ/2ε_{0})/mg

solve this eq. for σ.

Remember that length of the thread is not applicable in the question.

**Asked By: GAURAV MAHATE**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

initial energy of capacitor = 1/2 CV^{2}

final energy of capacitor = 1/2 KC (V/K)^{2}

difference ?

**Asked By: GAURAV MAHATE**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

initial energy of capacitor = 1/2 CV^{2}

final energy of capacitor = 1/2 KC (V/K)^{2}

difference ?

Calculate potential difference across a and b ,

**Asked By: AMIT DAS**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

since net potential = 0, answer will be 0 for the given diagram. By interchanging the poles of any one battery this question will become meannigfull so change it and apply Kirchhoff law.