Find maximum charge on ball and maximum current through inductor .

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An uncharged conducting ball of radius r is earthed through an inductor
L as shown in the figure. A horizontal electron beam swoops on the ball
from a great distance. The density of electrons in the incident beam is n
and all these electrons are moving with velocity u that is much smaller
than the speed of light so the relativistic effects can be neglected .
Find maximum charge on ball and maximum current through inductor .

asked Sep 4, 2018
retagged Sep 5, 2018

1 Answer

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Your method for solving the problem is correct, but you have made some mistakes.

First, you have assumed that the current $i$ through the inductor towards Earth equals $-\frac{dq}{dt}$ where $q$ is the charge on the sphere. This is not quite correct because charge is also arriving on the sphere from the electron beam at the constant rate $i_0$. So you should have $$\frac{dq}{dt}=i_0 -i$$

Then we can differentiate and substitute into your 1st equation : $$\frac{q}{C}=L\frac{di}{dt}=-L\frac{d(i_0-i)}{dt}$$ $$\frac{1}{C}\frac{dq}{dt}=\frac{1}{C}(i_0-i)=-L\frac{d^2(i_0-i)}{dt^2}$$ $$\frac{d^2(i_0-i)}{dt^2}+\frac{1}{LC}(i_0-i)=0$$ $$(i_0-i)=A\sin(\omega t+\phi)$$ where $\omega^2 =\frac{1}{LC}$. I have used $\sin$ but you can instead use $\cos$ then the value of $\phi$ is different.

$A, \phi$ are constants which are to be determined from the initial conditions. When $t=0$ there is no charge on the sphere and no current flowing through the inductor ($i=0$) and also no voltage across the inductor ($V_L=L\frac{di}{dt}=0$) therefore $$i_0=A\sin\phi$$ $$-\frac{di}{dt}=0=\omega A\cos\phi$$ $$\implies \phi=\frac{\pi}{2}, A=i_0$$ $$i=i_0(1-\cos\omega t)$$ We can see from this that the maximum current through the inductor is $2i_0$, which occurs when $\cos\omega t=-1$.

So what is $i_0$? This is where you make another mistake or two. The rate at which charge falls onto the sphere is $$i_0=\text{cross-section area x velocity of electrons x density of beam x electron charge }=-\pi R^2 une$$ (Comment : This is not realistic because as -ve charge accumulates on the sphere other electrons in the beam will be repelled from it, so $i_0$ will not be a constant, as assumed above, but will decrease and will be a function of $q$ and therefore of time $t$. However, I think we are expected to ignore this effect, which would make the problem vastly more complicated.)

What about maximum charge? We integrate the equation for $i(t)$ using the initial condition $q=0$ when $t=0$ : $$i(t)=i_0-\frac{dq}{dt}=i_0(1-\cos\omega t)$$ $$\frac{dq}{dt}=i_0\cos\omega t$$ $$q=\frac{i_0}{\omega}\sin\omega t$$ Maximum charge occurs when $\sin\omega t=\pm 1$ and is $$Q=\pm \frac{i_0}{\omega}=\pm \sqrt{LC}\pi R^2 une$$

answered Sep 8, 2018 by (26,660 points)
selected Sep 9, 2018 by koolman
@sammy_gerbil sorry Sr but how did you get $i=i_0(1-\cos\omega t)$? I mean, I have been trying to get it from the derivative of $(i_0-i)=A\sin(\omega t+\phi)$.
@JD_PM Apply the initial conditions $i=0$ and $\frac{di}{dt}=0$ at $t=0$, as shown in my answer.
How is it possible that at t=0 you get an equation where it does appear t?
I do not understand what you mean. What equation?
$i = i_o( 1 - cos\omega t)$
Substitute $\phi=\frac{\pi}{2}, A=i_0$ into $i_0-i=A\sin(\omega t+\phi)$. Does that answer your question? I do not understand what you mean by "How is it possible that at t=0 you get an equation where it does appear t?"
My bad Sr. I did not thought that $sin(\omega t + \pi/2) = cos(\omega t)$