Shells Losing Energy !

Electricity and Magnetism Level 4

An arrangement of two concentric conducting spherical shells charged initially such that the inner shell (radius = $a$ ) carries a net charge of $+Q$ while outer shell (radius $b$ ) carries net charge $0$ . If the switch " $S$ " is closed, how much electrostatic potential energy would the system lose?

\frac{Q^2}{4 \pi \epsilon_0 a}

\frac{Q^2}{4 \pi \epsilon_0 b}

\frac{Q^2}{8 \pi \epsilon_0 a}

\frac{Q^2 (b-a)}{4 \pi \epsilon_0 ab}

None of These

\frac{Q^2}{8 \pi \epsilon_0 b}

3 solutions

Rohit Gupta
May 29, 2015

Sir ,here,loss in Potential Energy is exactly same as self-energy of conducting sphere . Is there any relation between both of them??

Akhil Bansal - 5 years, 10 months ago

Nishant Rai
May 27, 2015

Before the switch is closed there is an induced charge of $-Q$ on the inner surface of the outer shell (radius $b$ ) and an induced charge of $+Q$ on it’s outer surface. Thus the total Potential Energy of the system is $\large U_i = \frac{Q^2 (b-a)}{8 \pi \epsilon_0 ab} + \frac{Q^2}{8 \pi \epsilon_0 b}$

whereas when the switch is closed the outer surface charge $+Q$ is absorbed by the ground and only the charge $-Q$ on the inner surface remains (no change for the inner sphere ), hence the potential energy of the system becomes $\large U_f = \frac{Q^2 (b-a)}{8 \pi \epsilon_0 ab}$ . Hence loss in potential energy is $\large \frac{Q^2}{8 \pi \epsilon_0 b}$

In the expression Ui shouldn"t be 8 instead of 4

Kyle Finch - 6 years ago

I can't understand the concept of positive charge being absorbed by the ground

Atul Solanki - 6 years ago

Refer to the concept of Earthing of Conductors -

Because when a body is earthed its potential become same as that of the earth. And since conventionally the earth is considered to be of zero potential hence the body is said to be of zero potential.

@Atul Solanki

Nishant Rai - 6 years ago

but how can positive charge go to earth

Atul Solanki - 6 years ago

@Atul Solanki – Electrons flow from the shell to earth. It's the same as positive charges flowing from earth to the shell, just like the conventional electric current

João Vitor Cordeiro de Brito - 6 years ago

Here,loss in Potential Energy is exactly same as self-energy of conducting sphere . Is there any relation between both of them??

Akhil Bansal - 5 years, 10 months ago

Nice solution

Satyam Tripathi - 4 years, 5 months ago

Can't understand why the potential energy before closing the switch is the Ui. Why it is not just the auto energy of the spherical shell of radius a?

$AE=\cfrac { { Q }^{ 2 } }{ 8\pi { \epsilon }_{ 0\quad \\ }a } \quad \\$

(I'm sorry my TeX isn't working properly above)

João Vitor Cordeiro de Brito - 6 years ago

Correct me if i am wrong,

What you have termed as AUTO ENERGY is what SELF POTENTIAL ENERGY of a body is.

Due to induction, the inner surface of the outer shell accumulates a charge of $-Q$ and the outer surface of the outer shell a charge of $+Q$ .

If you look closely, the outer surface of inner shell and inner surface of the outer shell makes a circular capacitor charged with charge $Q$ . So one has to include the energy stored in the capacitor while writing $U_i$ .

And self potential energy for the the outer surface of the outer shell is $\huge U_{self} = \frac{Q^2}{8 \pi \epsilon_0 b}$

@João Vitor Cordeiro de Brito

Nishant Rai - 6 years ago

Yes, I am really sorry, what I call "auto energy" is the self energy (Self is translated as Auto in portuguese and I forgot translating) Thanks.

Got it! You are right, they form a capacitor. I supposed that the radius b shell was sufficiently thin, and I neglected its energy. Thanks

João Vitor Cordeiro de Brito - 6 years ago

Sidharth Nair
May 21, 2018

On earthing the outer sphere a negative charge of -Q will appear on it. Hence its energy loss has to be ${kQ^2/2b}$ , where k is $1/4\pi\epsilon$

Shells Losing Energy !

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