Charge redistribution

Electricity and Magnetism Level 5

Consider a neutral conducting sphere of radius $R=1~\text{cm}$ . It is clear that if there are no other charges nearby, the electrostatic energy of the sphere is zero. However, if a charge $q$ is placed in the vicinity of the sphere, the charges on its surface redistribute and, as a consequence, the sphere acquires an electrostatic potential energy $\Delta{U}$ . Determine the interaction energy $\Delta{U}$ in Joules among the charges on the surface of the sphere if a charge $q=1~\mu \text{C}$ is placed at a distance $d=3~\text{cm}$ from its center.

Details and assumptions

$k=\frac{1}{4\pi \epsilon_{0}}= 9\times 10^{9}~\text{m/F}$

The answer is 0.00625.

2 solutions

Aniket Sanghi
Mar 7, 2017

I will 1st tell you my trick . I am a JEE ASPIRANT , So I prefer to derive short methods. :)

First Find the interaction potential energy of the interaction between the image charges and the charge we brought.

If it is like F. Then the interaction potential energy of the induced charges with itself comes out to be -F/2 .

And the total Interaction energy comes out to be F/2.

This can be easily proved , just find out the total work done on the charge by the external agent to bring it from infinity to that point . You will get total energy .

This will be F/2! Hence proved : ).

here when be bring a 1 micro columb charge near to the sphere. this charge will exert force on the free electrons of the sphere. these electrons will move in such a way that the net electric field at any point on the sphere is radial. when we bring a charged object in from of neural conducting object interesting things happen. interesting thing is in the begining the free electrons of the conductor start moving due to the electrostatic force. but after sometime. they settle on the surface of the sphere. they electrons move in such a way that the net electric field at any point on the surface has tangential component zero.

Srikanth Tupurani - 11 months, 1 week ago

A Former Brilliant Member
Mar 7, 2017

A no of methods can be used perhaps.I know 2-3.One involves integration,another simply placing the image charges and calculating their interaction energy(divide by 2 because of space symmetry). Another approach could be to bring the charge from infinity and calculating the work done gets stored as total electrostatic energy of the system )Finding the surface charge density from superposition and gauss theorem.2)The potential at any point on the sphere due to itself can be found out.And considering a ring and using and dq=sigma*area of ring.But the energy of interaction is (U/2).Since here we have considered the same interaction twice Courtesy And Solution Source @Spandan Senapati He is the one to solve it rightly before me (lost it bcoz of silly mistake.)

Thank you @shubham dhull

Spandan Senapati - 4 years, 3 months ago

The one involving the integration was a bit tedious but that was overcome by more efficient methods.As we could ring a symmetric element over which integration could be done....In a more weird system this is the approach(Standard☺☺)

Spandan Senapati - 4 years, 3 months ago

no problem bro. i never lie :P

A Former Brilliant Member - 4 years, 3 months ago

@A Former Brilliant Member – Good BTW which DAV you study??

Spandan Senapati - 4 years, 3 months ago

@Spandan Senapati – O.S.D.A.V. public school

A Former Brilliant Member - 4 years, 3 months ago

Charge redistribution

The answer is 0.00625.

2 solutions

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