Electric field on surface of earthed conductor

[CAF123]

Homework Statement

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.

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.

 

[rude man]

Homework Statement

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.
wtf? EVERY point on the surface of the sphere is "directly above the center of the sphere"!

The Attempt at a Solution

So q is a distance $\sqrt{a^2 + d^2}$ from the centre.

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?

 

[SammyS]

Homework Statement

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 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.

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.

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.

 

[CAF123]

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.

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

Yes, sorry should have made that clear.

If the charges are located along the x-axis, and point P is on the y-axis, at y=a,

Yes

This will not show directly that the other components of E are zero.

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.

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

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

 

[SammyS]

...

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

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.

 

[CAF123]

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

 

[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.

 

[CAF123]

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

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.

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.

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.

 

[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.

 

[CAF123]

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

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.

 

[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.

 

[SammyS]

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.

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.

 

[CAF123]

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)?

Take one point as (-d,0) and the other (0,a). Then the vector connecting these is $d \hat{i} + a \hat{j}$

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 $\hat{r} = \frac{d}{\sqrt{d^2 + a^2}} \hat{i} + \frac{a}{\sqrt{d^2 + a^2}} \hat{j}$

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

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

 

[CAF123]

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.

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.

 

[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.

 

[SammyS]

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.

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.

 

[SammyS]

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.

They don't cancel if you take the derivative.

 

[CAF123]

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.

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)?

 

[rude man]

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

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.

 

[SammyS]

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)?

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 .

 

[CAF123]

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.

Thanks, the only problem I have is that the x components aren't cancelling and they should since the E field is perpendicular to the conductor and so it is only in -y direction.

 

[rude man]

Thanks, the only problem I have is that the x components aren't cancelling and they should since the E field is perpendicular to the conductor and so it is only in -y direction.

Oh, but it does! If you substitute your b and q₁ expressions in terms of q, a and d, you will find that E_x = 0. In other words, E = ( ) j only. As required of course.

 

[SammyS]

Oh, but it does! If you substitute your b and q₁ expressions in terms of q, a and d, you will find that E_x = 0. In other words, E = ( ) j only. As required of course.

For what it's worth, I agree!

 

[CAF123]

For what it's worth, I agree!

Oh I made an error before I see it now. Thanks rude man and SammyS!