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Introduction

Six point charges of magnitude $Q_0$ is evenly distributed on a sphere with radius $a$ and center in the origin. The six point charges have the cartesian coordinates:

$q_1=(a,0,0) \: \: \: \: q_2=(-a,0,0)$

$q_3=(0,a,0) \: \: \: \: q_4=(0,-a,0)$

$q_5=(0,0,a) \: \: \: \: q_6=(0,0,-a)$

The electric field caused by these six point charges is denoted $\mathbf{E_0}$.

A point charge of magnitude $-6Q_0$ is placed in the origin of that same coordinate system. The electric field caused by this point charge is denoted $\mathbf{E_1}$.

Problem

A conducting spherical shell with radius $\frac{a}{2}$ is now placed with its centre in the origin. The seven point charges are still in place.

Determine the total electric field $\mathbf{E=E_0+E_1}$ inside the region surrounded by the conductor.

Also determine the surface charge density $\rho$ on the inner surface of the conductor.

My attempt

Since the conductor is closed, the enclosed by the conductor is "protected" from any field there must be outside the conductor. Therefore, we have that $$\mathbf{E=E_1} $$

However, I'm really uncertain how I should determine the surface charge density, $\rho$. Can someone help me with this?

I really need help.

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  • $\begingroup$ You cannot use gauss law here since you don't have symmetry, you wrote that there is 6 point charges on the edge of the sphere, you didn't say there is a surface density $\sigma$ $\endgroup$
    – Sagigever
    Sep 20, 2020 at 14:02
  • $\begingroup$ Correct, the six point charges are on a sphere and not on the conductor surface. Hmm, if I can't use Gauss' Law what method could I then use? @Sagigever $\endgroup$
    – Carl
    Sep 20, 2020 at 14:15
  • $\begingroup$ I would consider the super position principle, calculate the electric field of each point charge by the formula $\vec{E}=\frac{kq(\vec{r}-\vec{{r'}})}{|(\vec{r}-\vec{{r'}}|)^3}$ $\endgroup$
    – Sagigever
    Sep 20, 2020 at 15:08

2 Answers 2

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We need to have a considerable thickness of the material so that we can neglect the discreteness of charge. Let us consider the shell to have a thickness $t$ which is extremely small.

Now, assume the point $D$ inside the meat of the conductor as shown:

enter image description here

The field inside the meat of conductor is always zero. The charge on the outer surface will distribute themselves to shield the effect of the outer $6$ charges.

So, the charge on inner surface must distribute itself such that the Electric field at $D$ is zero. So it would try to shield the effect of the $-6Q_0$.

The only way it can be done is that the inner surface should behave as a sphere having a uniform charge of $+6Q_0$.

Now, you can calculate $\rho$ easily. Cheers :)

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  • $\begingroup$ So, is it just $\rho = \frac{6Q_0}{4\pi \cdot (\frac{a}{2})^2}$? Thanks for the answer, by the way. $\endgroup$
    – Carl
    Sep 21, 2020 at 7:47
  • $\begingroup$ Yes... just that. I hope my answer gets accepted :P $\endgroup$ Sep 21, 2020 at 8:03
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    $\begingroup$ Haha, there you go, you’ve earned it. Thanks for helping a man in need :) $\endgroup$
    – Carl
    Sep 21, 2020 at 8:05
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Because it is a conducting sphere one can implicitly say that any field/charges outside the sphere cannot affect the field inside as a natural result of a conducting boundary. Similarly any inside charge cannot affect the field outside the conductor. I know this seems a bit weird at first but that is one of the properties of a perfect conductor. So you can use Gauss' law to calculate the field on an inside point up to a/2 distance from the origin. For the surface charge density on the inside of the conductor, calculate the charge that will get accumulated there because of the charge present at the origin and it comes out to be equal to the said charge. Then for average density divide that by the surface area of the sphere. Can't say about the density function though...

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