12 - Capacitors Questions Answers

Physics >> Electricity >> Capacitors Medical Exam

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 )q2/(C/2)   ie 1/4 q2/C

BUT Sir the ans given is=( 1/2)q2/C Plzz do me a little favour by solving this query

Asked By: SARIKA SHARMA
is this question helpfull: 7 0 read solutions ( 1 ) | submit your answer
Joshi sir comment

(1/2 )q2/(C/2) = q2/C not 1/4 q2/C

so work done = new energy - previous energy

Physics >> Electricity >> Capacitors Medical Exam

a II plate capacitor has an electric field of 105 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
is this question helpfull: 4 0 submit your answer
Joshi sir comment

F = σq/2ε0 = Eq/2 = 105*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
is this question helpfull: 5 1 submit your answer
Joshi sir comment

 

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

One end of a 10cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate having a mass 10g and a charge of 4*10^-6 c . In equlibrium the thread makes an angle of 60 deg. With he vertical.find the surface charge density on plate.
Asked By: GAURAV MAHATE
is this question helpfull: 2 1 read solutions ( 1 ) | submit your answer
Joshi sir comment

according to the diagram 

tanθ = qE/mg = q(σ/2ε0)/mg

solve this eq. for σ.

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

there is slight mistake in last line of previous question.it was find the work done on the system in the process of inserting the slab.
Asked By: GAURAV MAHATE
is this question helpfull: 4 1 submit your answer
Joshi sir comment

 

initial energy of capacitor = 1/2 CV2

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

difference ?

a parallel plate capacitor of plate A and plate seperation d is charged to a potential dif. V and then the battery is disconnected .a slab of dielectric constant K is then inserted between the plates of capacitor so as tofill the space bet. The plates .find the work done in this process.
Asked By: GAURAV MAHATE
is this question helpfull: 3 0 read solutions ( 1 ) | submit your answer
Joshi sir comment

initial energy of capacitor = 1/2 CV2

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

difference ?

Calculate potential difference across a and b ,

Asked By: AMIT DAS
is this question helpfull: 3 0 submit your answer
Joshi sir comment

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.

a 24.6 mev alpha particle moves towards gold nucleus[atomic no 79] calculate the distance of closest approach of alpha particle

Asked By: SHUBHAM SHARMA
is this question helpfull: 3 0 submit your answer
Joshi sir comment

upto closest approach total KE of α particle will become PE

so use the formula KE = 9*109 * Ze2e/r

here KE = 24.6 mev   convert it into joule first

solve equation for r 

a capacitor consists of two metal plates each 10cm by 20cm ,they are seperated by a 2 mm thick insulator with dielectric constant 4.1and dielectric strength 6*10to power7 v/m.what minimum voltage will cause spark in capacitor.
Asked By: GAURAV MAHATE
is this question helpfull: 3 3 submit your answer
Joshi sir comment

V = Ed = 6* 107*0.002

Define Potential Energy?

Asked By: SIDDHANT KAPOOR
is this question helpfull: 3 0 submit your answer
Joshi sir comment

Potential energy is the energy of a particle due to its position relative to the reference point taken.

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