Register to reply 
Surface charge induced in conductor by point charge 
Share this thread: 
#1
Nov1908, 04:31 AM

P: 527

1. The problem statement, all variables and given/known data
Find the surface charge induced on an infinite conducting plate by a point charge +q at a distance a from the plate. 3. The attempt at a solution As you probably know, this problem can be solved using the wellknown fact that the induced electric field is equivalent to replacing the conductor by a negative point charge q placed diametrically oposite to +q. I am trying to solve the problem without using this fact. Can you spot the error in this logic? The electric field at a point on the surface due to the point charge +q is given by [itex]\frac{q}{4\pi\epsilon_0 }\frac{a}{r^3}[/itex]. Since the plate is conducting, the surface electric field is [itex]\frac{\sigma}{\epsilon_0}[/itex]. Application of Gauss's law to a pillbox of negligible height and area A, gives [itex]\oint \vec{E} \cdot d\vec{a} = \frac{Q_\mathrm{encl}}{\epsilon_0}[/itex] [itex]\frac{q}{4\pi\epsilon_0}\frac{a}{r^3} A  \frac{\sigma}{\epsilon_0}A = \frac{\sigma}{\epsilon_0}A \implies[/itex] [itex]\sigma = \frac{qa}{8\pi\epsilon_0 r^3}[/itex]. Correct answer: [itex]\sigma = \frac{qa}{2\pi\epsilon_0 r^3}[/itex]. Edit: Corrected typos. 


#2
Nov1908, 04:51 AM

HW Helper
P: 5,003

I see three problems here...
(1)the field due to the point charge is not [tex]\frac{q}{4\pi\epsilon_0 }\frac{a}{r^3}[/tex] (in fact, this isn't even the correct magnitude for the field) (2)the electric field at the surface to due the induced surface charge is really undefined....on one side of the plane, it points in one direction with a magnitude [tex]\frac{\sigma}{2\epsilon_0}[/tex] and on the other side, it has the same magnitude, but points in the opposite direction!...So when you talk about the field at the surface...are you talking about the field above the surface? the field below? the average field?.....The answer is that the field at the surface is really infinite (undefined)... (3) the third problem is that while the field on either side of the plane due to just the induced charge is uniform over your gaussian surface, the field due to the point charge is not and so [tex]\oint \vec{E}_{\text{point charge}} \cdot d\vec{a} \neq \frac{q}{4\pi\epsilon_0 }\frac{a}{r^3}A[/tex] PS if the conducting plane is the xy plane, then the correct surface charge density is [tex]\sigma(x,y) = \frac{qa}{2\pi (x^2+y^2+a^2)^{3/2}}\neq \frac{qa}{2\pi\epsilon_0 r^3}[/tex] 


#3
Nov1908, 05:28 AM

P: 527

Thanks for your reply gabbagabbahey.
None of these are actually problems. (1) The expression I gave is the magnitude of the perpendicular component of the electric field at the surface of the plate due to the positive point charge (the horizontal component can be ignored by making the height of the pillbox vanishingly small). r is the distance between the charge and the point on the plate. (2) The electric field inside the matirial is zero (since it is a conductor). The expression for the electric field on the exterior is sigma/epsilon_0, not sigma/2epsilon_0. It is welldefined. (3) This is also not a problem: just take the area intercepted by the Gaussian surface to be sufficiently small that the perpendicular component of the electric field (and the surface charge density) can be considered constant. 


#4
Nov1908, 05:45 AM

HW Helper
P: 5,003

Surface charge induced in conductor by point charge
EDIT shouldn't the perpendicular component be [tex]\frac{q}{4\pi\epsilon_0 }\frac{a}{r^3} \cos \theta[/tex] where theta is the angle between the vector from the point charge the the point on the plane, and the vector from the point charge to the center of the plane? The plane is considered to be a single infinitesimally thin surface, so it makes no sense to say the field inside the material is zero because the material in this case is infinitesimally thin. You have a single surface; not two distinct surfaces with material in between! The electric field at the surface is undefined in this case, because it simultaneously points in two directions! The electric field on either side of the surface, is however welldefined and it has a magnitude of [tex]\frac{\sigma}{2\epsilon_0}[/tex] (you can check your text if you don't believe me about the factor of 2) and points downward below the surface, and upward above it. When you use your Gaussian pillbox, you have two surfaces (neglecting the sides of the box), and you need to be careful to calculate the flux through both surfaces....as long as your box is of finite thickness, the fact that the field at the surface is undefined won't be a problem.....just make sure to place your pillbox so that its top is above the plane, and its bottom is below... 


#5
Nov1908, 05:54 AM

P: 527

I don't know why you think the conductor should have negligible thickness, nor why the charge density should be sigma/2epsilon_0.
You can use a Gaussian surface to easily show it is sigma/epsilon_0. The result you quoted would hold for a nonconducting slab with surface charge e.g. 


#6
Nov1908, 06:04 AM

HW Helper
P: 5,003

I was thinking of an infinite conducting "plane", but now rereading the question I see you have an infinite conducting "plate"...which can have a nonzero thickness. ...In this case, then yes the field is just sigma/epsilon_0.
However, the perpendicular component of the field due to the charge is still incorrect....you need a cos(theta) there. 


#7
Nov1908, 06:05 AM

P: 527

No. because [itex]\cos\theta = a/r[/itex].



#8
Nov1908, 06:06 AM

P: 527

[itex]\therefore E_\perp = E\cos\theta = (q/4\pi\epsilon_0 r^2)(a/r) = qa/(4\pi\epsilon r^3)[/itex]



#10
Nov1908, 06:22 AM

P: 527

So, I wonder why I'm getting the wrong answer.
Could it have something to do with the fact that the problem must be solved selfconsistently? 


#11
Nov1908, 06:43 AM

HW Helper
P: 5,003

Are you sure you are looking for the case of a conducting plate with some finite thickness?.....the method of images doesn't really work in that case....
In the case of a conducting "plane" on the other hand, it does and it gives [tex]\sigma = \frac{qa}{2\pi\epsilon_0 r^3}[/tex] Anyways, in the case of the thick conducting slab, the field due to the induced surface charge points upwards so [tex]\oint \vec{E}_{\text{surface}} \cdot \vec{da}=\frac{+\sigma A}{\epsilon_0} \neq \frac{\sigma A}{\epsilon_0}[/tex] but you also have a contribution due to the other surface of the slab (the one not enclosed by the gaussian pillbox) since it creates a nonzero field too! In the case of the conducting "plane" on the other hand, there is only one surface charge density to consider,and it creates a field of strength sigma/2epsilon_0 , and so [tex]\oint \vec{E}_{\text{surface}} \cdot \vec{da}=\frac{+\sigma A}{2\epsilon_0}[/tex]...which will yield your desired answer. 


#12
Nov1908, 06:54 AM

P: 527

Nah, images should work fine for an infinitethickness slab. In fact, that's the original idea it was proposed for I believe.
Also, the field due to the induced charge doesn't point up (remember a +ve point charge induces a ve surface charge). The equations I gave in my original post stand. 


#13
Nov1908, 06:56 AM

P: 527

Oh, and the contribution from ``the other side'' is irrelevant because the sheet is infinite so they will never come in contact. Alternatively, just send the other side away to infinity by considering a thick slab.



#14
Nov1908, 06:57 AM

HW Helper
P: 5,003

Where would you put the image charge for an infinite thickness slab???....you can't have an image charge inside the slab, because it would create a nonzero field there and so it wouldn't satisfy the boundary condition that the field is zero everywhere inside the conductor!



#15
Nov1908, 06:59 AM

HW Helper
P: 5,003

Using the Gaussian method for the slab of infinite thickness, you would have to consider a slab of finite thickness 'd' and then take the limit as d approaches infinity....the contribution due to the other surface will make a difference.
If you don't include the term due to the other surface, you have: [tex]\frac{q}{4\pi\epsilon_0}\frac{a}{r^3} A + \frac{\sigma}{\epsilon_0}A = \frac{\sigma}{\epsilon_0}A \Rightarrow \frac{q}{4\pi\epsilon_0}\frac{a}{r^3} A=0[/tex] which is a contradiction for a nonzero A. 


#16
Nov1908, 07:20 AM

HW Helper
P: 5,003




#17
Nov1908, 06:30 PM

P: 527

Let me see,
Gauss' law is [itex]\frac{qa}{4\pi\epsilon_0 r^3} A  \frac{\sigma}{\epsilon_0} A = \frac{\sigma}{\epsilon_0} A[/itex]. So it does indeed cancel off. Hmmm, interesting. 


#18
Nov1908, 09:28 PM

HW Helper
P: 5,003

Yes, and in order to account for that apparent contradiction, in the case of the thick slab, you need to add the flux due to the other surface charge....
However, you may want to reread the original question, because if the answer is supposed to what you said it was, the problem must actually be about an infinitesimally thin slab (aka a plane); and the field due to that is [tex]\frac{\sigma}{2\epsilon_0}[/tex]....which will give you your desired answer. 


Register to reply 
Related Discussions  
Point charge at the center of a spherical conductor  Introductory Physics Homework  17  
Induced charge on a grounded conductor  Advanced Physics Homework  4  
Charge induced on conductor  Classical Physics  0  
Finding the charge on the outer surface of the conductor  Introductory Physics Homework  3  
Surface Charge of Conductor  Classical Physics  4 