# Electric field on surface of earthed conductor

1. Feb 9, 2013

### CAF123

1. The problem statement, all variables and given/known data

A charge q is a distance d from the centre of an earthed conducting sphere of radius a
(d > a).
Given that the image charge for this system is is q1 located a distance b from the centre of the sphere where $q_1 = -\frac{aq}{d},\,\,b = \frac{a^2}{d}$,
calculate an expression for the E field at point P on the surface of the sphere, directly above the centre of the sphere.

3. The attempt at a solution

So q is a distance $\sqrt{a^2 + d^2}$ from the centre. q1 is a distance $\sqrt{b^2 + a^2} = \sqrt{(a^4/d^2) + a^2}$away. Put these into the expression for potential and sum them together gives: $$V_T = V_q + V_{q_1} = \frac{q}{4 \pi \epsilon_o} \left[\frac{1}{\sqrt{a^2 +d^2}} - \frac{1}{\sqrt{a^2 + d^2}}\right] = 0$$

The E field will be perpendicular to the surface equipotential, and so will point in -y direction assuming positive y is defined vertically upwards. I know in general E = -∇V, but the above is zero (as expected). Should I instead just consider the total electric field using the value of b given?

This would give: $$E_T = \frac{q}{4 \pi \epsilon_o} \left[\frac{1}{a^2 +d^2}\underline{r} - \frac{1}{(a^4/d^2) + a^2}\underline{r'}\right] ,$$ where $\underline{r}$ is the direction of the E field vector (compts in both x and y) for q and similarly for $\underline{r'}$ for the E field direction by $q_1$.

I think my next step would be to write the vector r decomposed into x and y and then see that the x components of the sum of the two E field vectors should cancel, although I am not sure how to do this.

Last edited: Feb 9, 2013
2. Feb 9, 2013

### rude man

No, that would be d.

The wording of the problem being - er - problematical, are you just trying to find the potential at the point on the surface on a line between q and -q1? Maybe they want you to find the field (and so the potential) at every point on the surface of the sphere, or on any point along a great circle joined by that line ... hard to say.

Anyone else comment?

3. Feb 9, 2013

### SammyS

Staff Emeritus
There appear to be some contradictions in what you have written.

The statement of the problem says that
"…charge q is a distance d from the centre of an earthed conducting sphere of radius a…"​
and
"…the image charge for this system is q1 located a distance b from the centre of the sphere…" .​
Then at the beginning of your solution, you state
"So q is a distance $\sqrt{a^2 + d^2}$ from the centre. q1 is a distance $\sqrt{b^2 + a^2} = \sqrt{(a^4/d^2) + a^2}$away."​
Perhaps point P is located at a position on the surface of the sphere such that the two charges are those distances away from point P.

Regarding the use of the potential to find E at the surface, you haven't actually stated where the charges are located in your coordinate system, other than their location relative to the center (centre) of the sphere, which I assume is at the origin.

If the charges are located along the x-axis, and point P is on the y-axis, at y=a, then you could replace a with y in your expression for the potential, using that to find the y-component of E along the y-axis. (This expression for the potential is zero at y=a, but it's good for any y≥a .) This will not show directly that the other components of E are zero.

To get all the components of E from the potential, you need to find the potential at an arbitrary point external to the sphere.

4. Feb 9, 2013

### CAF123

Yes, that is right, I made a mistake. The distances given are the distances of the charges from point P.

Yes, sorry should have made that clear.

Yes

I suppose I woudn't really need to show this - from the fact that E-field lines strike the conductor perpendicular, I know in advance that it will all be in -y.

If I wanted to show that the x compts cancel at point P, how would this help?

5. Feb 9, 2013

### SammyS

Staff Emeritus
I suppose that it wouldn't help. As you stated, you know the E field is perpendicular the conductor at its surface, so all you need is the y-component to find the E field at point P.

6. Feb 10, 2013

### CAF123

So having attained my expression for the E field in my opening post, what should I do now?

7. Feb 10, 2013

### rude man

If we now all agree that:
xyz coord. system origin is at center of sphere
q sits at x = + d, d > a
-q' sits at x = -b, b < a
P sits at (0,a,0) i.e. on the +y axis at y = a

THEN we can continue.
So you can let
r1 = distance from q to P
r2 = distance from -q' to P
So V = kq/r1 - kq'/r2, k = 9e9 SI

So now E = - grad V
So you need to compute the partial derivatives inherent in grad V, or

grad V = ∂V/∂x i + ∂V/∂y j + ∂V/∂z k, where i, j, k = unit vectors.
So now you need to determine ∂(1/r1)/∂x etc. etc.
All just analytic geometry.

8. Feb 10, 2013

### CAF123

If r1 and r2 are distances of the charges to point P then putting these distances into the eqn obtained via Pythagoras gives zero. (the charges are placed such that V =0 on surface of sphere). So I don't have an expression to work with.

Instead I just found the expression for the E field directly. I have this in term of vector r and vector r' for charges q and q'. I think I need to express this in terms of x and y.

9. Feb 10, 2013

### rude man

Looking at this some more, unless you're directed to use potentials, I would just compute the vector E fields directly. Much easier.

10. Feb 10, 2013

### CAF123

That is what I did, but I am not sure what to do from here. The expression I have is given in the OP but it includes $\underline{r}$ and $\underline{r'}$, and somehow I need to reduce this to something in $\hat{y}$ only.

11. Feb 10, 2013

### rude man

Well, since by symmetry the z component of E = 0 for q and q',

So for q, |E| = kq/(d^2 + a^2) and E = kq/(d^2 + a^2) with direction from q to P.

If I give you two points (x1, y1) and (x2, y2) what is the vector going from (x1, y1) to (x2, y2)? And what is the normalized vector (magnitude = 1)? And do you see that the E vector will be [kq/(d^2 + a^2)]*(normalized vector) from q to P?

So do that for q and q' (remember, q' < 0) to get the E field due to both charges.

12. Feb 10, 2013

### SammyS

Staff Emeritus
I mentioned this in Post # 3, but not with these details.

Replacing a with y in your expression for the potential gives an expression for the potential that's valid along the y-axis from the exterior surface of the sphere & outward. (Sure, ir evaluates to zero at y=a.) However in doing this you should only replace a in the quantities that represent from the charges. Don't replace the a which is involved in calculating q1 or calculating b.

$\displaystyle V(x=0,\,y\ge a,\,z=0)= \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{\sqrt{y^2 +d^2}} + \frac{q_1}{\sqrt{b^2 + y^2}}\right)$

You can replace q1 with -aq/d and b with a2/d, either before or after taking the derivative with respect to y.

13. Feb 10, 2013

### CAF123

Take one point as (-d,0) and the other (0,a). Then the vector connecting these is $d \hat{i} + a \hat{j}$
So $\hat{r} = \frac{d}{\sqrt{d^2 + a^2}} \hat{i} + \frac{a}{\sqrt{d^2 + a^2}} \hat{j}$
q' is along the x axis, at position (-b,0) and consider P(0,a)
Similarly, for q', $\hat{r'} = \frac{b}{\sqrt{b^2 + a^2}} \hat{i} + \frac{a}{\sqrt{a^2 + b^2}} \hat{j}$

This gives: $$\underline{E} = \frac{1}{4 \pi \epsilon_o} \left(\frac{q}{d^2 + a^2}\left(\frac{d}{\sqrt{d^2 + a^2}}\hat{i} + \frac{a}{\sqrt{d^2 + a^2}} \hat{j}\right) +\frac{q'}{b^2 + a^2}\left(\frac{b}{\sqrt{b^2 + a^2}} \hat{i} + \frac{a}{\sqrt{a^2 + b^2}} \hat{j}\right) \right).$$

14. Feb 10, 2013

### CAF123

Is there a way to use potentials to get the E field at point P? Since it evaluates to zero at P (the point we are interested in), how would evaluating at some arbritary point on y help?

If I substitute the conditions given in the question for q' and b, then the expression you wrote will cancel to zero (as expected).
Thanks.

15. Feb 10, 2013

### CAF123

When I multiply out my expression given in my last post, I'd hope that the x components would cancel after I had substituted in the conditions for q' and b. But they don't.

16. Feb 10, 2013

### SammyS

Staff Emeritus
Any function defined at a point evaluates to some number at that point. By writing V as a function of y, we can take it's derivative. Since point P is on the y-axis, we can evaluate the derivative at point P.

17. Feb 10, 2013

### SammyS

Staff Emeritus
They don't cancel if you take the derivative.

18. Feb 10, 2013

### CAF123

I see. So I subbed things in too early. The expression I get is:$$E(y=a) = \frac{1}{4 \pi \epsilon_o} \left[ \frac{-q}{(a^2 + d^2)^{3/2}} + \frac{aq}{d(a^2 + a^4/d^2)^{3/2}}\right]$$ Is this the final answer (i.e is there any check I can do to verify it)?

19. Feb 10, 2013

### rude man

This now looks 100% correct. You should collect all the i and j terms into one E expression: E = ( )i + ( )j. And factor out the (d^2 + a^2)3/2 and (b^2 + a^2)3/2 terms.

Also remember that q1 < 0.

20. Feb 10, 2013

### SammyS

Staff Emeritus
Right off hand, I don't know of an easy way to check it.

It looks like it all needs to be multiplied by -a .