Electric field on surface of earthed conductor

In summary, we are given a charge q located a distance d from the center of an earthed conducting sphere of radius a. The image charge for this system, q1, is located a distance b from the center of the sphere where ##q_1 = -\frac{aq}{d},\,\,b = \frac{a^2}{d}##. We are asked to calculate an expression for the electric field at point P, which is on the surface of the sphere directly above the center. Using the expression for potential, we can sum the potential of q and q1 to get a total potential of 0 at point P. This means the electric field will be perpendicular to the surface of the sphere and point
  • #1
CAF123
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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.
 
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  • #2
CAF123 said:

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?
 
  • #3
CAF123 said:

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.
 
  • #4
SammyS said:
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?
 
  • #5
CAF123 said:
...

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.
 
  • #6
So having attained my expression for the E field in my opening post, what should I do now?
 
  • #7
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
rude man said:
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.
 
  • #9
Looking at this some more, unless you're directed to use potentials, I would just compute the vector E fields directly. Much easier.
 
  • #10
rude man said:
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.
 
  • #11
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
CAF123 said:
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.

[itex]\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)[/itex]

You can replace q1 with -aq/d and b with a2/d, either before or after taking the derivative with respect to y.
 
  • #13
rude man said:
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: $$\underline{E} = \frac{1}{4 \pi \epsilon_o} \left(\frac{q}{d^2 + a^2}\left(\frac{d}{\sqrt{d^2 + a^2}}\hat{i} + \frac{a}{\sqrt{d^2 + a^2}} \hat{j}\right) +\frac{q'}{b^2 + a^2}\left(\frac{b}{\sqrt{b^2 + a^2}} \hat{i} + \frac{a}{\sqrt{a^2 + b^2}} \hat{j}\right) \right).$$
 
  • #14
SammyS said:
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.

[itex]\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)[/itex]

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.
 
  • #15
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
CAF123 said:
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.
 
  • #17
CAF123 said:
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.
 
  • #18
SammyS said:
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)?
 
  • #19
CAF123 said:
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.
 
  • #20
CAF123 said:
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 .
 
  • #21
rude man said:
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.
 
  • #22
CAF123 said:
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 Ex = 0. In other words, E = ( ) j only. As required of course.
 
  • #23
rude man said:
Oh, but it does! If you substitute your b and q1 expressions in terms of q, a and d, you will find that Ex = 0. In other words, E = ( ) j only. As required of course.
For what it's worth, I agree!
 
  • #24
SammyS said:
For what it's worth, I agree!
Oh I made an error before I see it now. Thanks rude man and SammyS!
 

FAQ: Electric field on surface of earthed conductor

1. What is an electric field on the surface of an earthed conductor?

An electric field on the surface of an earthed conductor refers to the distribution of electric charges on the surface of a conductor that is connected to the ground. It is created when an electric current flows through the conductor and is equal in magnitude and opposite in direction to the charges on the surface of the conductor.

2. How is the electric field on the surface of an earthed conductor calculated?

The electric field on the surface of an earthed conductor can be calculated by dividing the surface charge density by the permittivity of free space. It can also be calculated by multiplying the electric potential by the gradient of the potential at the surface of the conductor.

3. What factors affect the strength of the electric field on the surface of an earthed conductor?

The strength of the electric field on the surface of an earthed conductor is affected by the amount of charge on the surface, the distance from the surface, and the presence of other conductors or insulators nearby. The shape and size of the conductor can also impact the electric field.

4. How does the electric field on the surface of an earthed conductor impact the surrounding environment?

The electric field on the surface of an earthed conductor can cause nearby objects to become charged and attract or repel each other. It can also induce currents in nearby conductors or disrupt electronic devices. Additionally, the electric field can be influenced by the environment, such as humidity and temperature.

5. How is the electric field on the surface of an earthed conductor used in practical applications?

The electric field on the surface of an earthed conductor is used in various applications such as lightning protection systems, grounding systems for electrical equipment, and in the design of high voltage power lines. It is also utilized in the study of electrostatics and for measuring the strength of electric fields in a given area.

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