4
$\begingroup$

Consider a thin spherical conducting shell of radius $R$, which carries a total charge $Q$ on its surface. Two point charges $Q$ and $2Q$ are at A and B respectively as shown in the figure ($C$ is the center of the shell).

enter image description here

If now the shell is earthed. How much total charge will flow into the earth ?

My attempt: Initially, the inner surface of the sphere will have a total charge $-Q$ distributed non-uniformly. Hence the outer surface will have a total charge $2Q$ (including the induced charges, whose sum amounts to zero) distributed non-uniformly.

After earthing, let the charge on the outer surface be $Q_1$. If $\vec{E}=0$ inside the shell, we get $V_{centre}= V_{shell}=0$. Putting the values with $k=\frac{1}{4π\epsilon_0}$, one gets $$\frac{kQ}{R/2}+\frac{k(2Q)}{2R}+\frac{k(-Q)}{R}+\frac{kQ_1}{R}=0.$$ (Since every point on the shell is equidistant from the centre , the charge distribution doesn't matter while calculating $V_{centre}$).

The above equation results in $Q_1=-2Q$, which further implies that $4Q$ charge flowed from the shell to the earth.

But the answer given is $3Q$. This is contradicting the fact that $\vec{E}=0$ inside the shell, which I can't understand since in my opinion existence of an electric field inside the shell will result in the movement of the charge inside, causing instability in the overall configuration. Please clear any misconceptions you find in this solution.

$\endgroup$
1
  • $\begingroup$ The R in your equation has not been defined. $\endgroup$
    – R.W. Bird
    Aug 3 '20 at 15:04
3
$\begingroup$

Your mistake: you assumed that V inside the cavity is zero.V is zero only in the material of the shell,not in the cavity as electric field can exist inside the cavity but not in the material of shell.Therefore you have to ensure that the potential of outer surface of shell is zero. There will be no potential at the outer surface of shell due to the charges Q at a distance r/2 from center of shell and the charge induced on inner surface of cavity as electric field lines cannot penetrate through the material of shell and hence the potential will only be due to the charge on outer surface of shell and the charge 2Q.The potential due to them both will be equal their potential at center of cavity as their combined electric field will terminate at outer shell surface.Hence, kQ1/R+k(2Q)/2r=0 which gives Q1=-Q(Q1 is final charge on outer surface of shell)but initially it had 2Q charge therefore 3Q charge flowed to earth

$\endgroup$
5
  • $\begingroup$ Thanks a lot! I understood that the charges $2Q$ outside and the charge $Q_1$ on the outer surface contribute zero electric field inside the sphere, collectively. So a net electric field exists inside the sphere .Now the electric field lines inside, cannot penetrate the inner wall of the conductor , which seems to conclude that the potential due to the charges $Q$ and $-Q$ , at the inner wall must be equal to that on the outer surface , which further implies that their combined potential at the inner surface must be 0 , for $V_{shell}$ to be 0 ( which I can't understand ). $\endgroup$
    – ash07
    Aug 3 '20 at 7:52
  • 1
    $\begingroup$ ok so you want to know the reason for the potential at inner surface of shell due to the charges Q and -Q to be zero.Go back to the definition of potential.It is the work done in bringing a unit positive charge from inifinity to the inner surface of shell in your problem.Notice that no field lines from Q and -Q can ever reach outside the inner surface of shell,hence you will perform no work against the electric field lines of Q and -Q in bringing a charge from infinity to that point.Hence their combined potential at the inner surface of shell is zero. $\endgroup$ Aug 3 '20 at 8:31
  • $\begingroup$ Thanks , All this time I was more focused on the $"\Sigma\frac{kq}{r}"$ definition of the electric potential . Now one last thing I want to make sure is that ,wouldn't the electric field inside will result in the movement of the charge $Q$ ? If it does , wouldn't it affect the stability of the system? $\endgroup$
    – ash07
    Aug 3 '20 at 9:11
  • 1
    $\begingroup$ physics.stackexchange.com/questions/107460/…. Read the second answer to this question to clear your doubt. $\endgroup$ Aug 3 '20 at 9:50
  • 1
    $\begingroup$ to ensure stability of the charge,they usually mention that the charge is fixed $\endgroup$ Aug 3 '20 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.