Point charge near charged plate

[swinfen]

Homework Statement

A conducting plate has a total charge of 10.0 C on the two sides of the plate of area 10.0 m2. What is the magnitude of the electric field on a charge of 3.0 mC only 2.0 nm from the surface?

Relevant Equations

E = σ/2e0

I thought the equation listed should be used, with the 'charge density' determined by the point charge multiplied by the area of the plate, but not sure if that makes sense.

 

[BvU]

Hi again,

This exercise is badly composed:

Your equation does not contain the magnitude of the test charge, so: no multiplication to calculate the electric field above a uniformly charged large conducting plate, leading to $E = \displaystyle {\sigma\over 2\varepsilon_0}$ and all you would have to do is calculate $\sigma$ .

However:

The 3 mC (test) charge will contribute to the electric field and change the surface charge distribution, in such a way that the electric field lines will be perpendicular to the plate. This is a more complicated situation that requires more advanced calculation, which was probably not intended by the exercise composer (unless this was already treated in your class, which I doubt)

$\$

 

[swinfen]

Thanks again BvU. That's kind of what I thought - that this seems like it might be more complex that the extent of the course (AP Physics 1 - which is pre-cal). I didn't calculate σ based on the total charge (I think incorrectly using the point charge), so if I were to ignore the point charge magnitude (as in 'it's tiny'), perhaps it's simply σ = q/A = 10/10 = 1 - or should it be 10/20 = 1/2 (as the charge is distributed across both sides of the plate - but if that were the case, wouldn't the electric field applied to the point charge on only be applicable from one side of the plate?...

 

[swinfen]

Thinking about it - I guess the 'charge density' must be the overall charge divided by the overall area - so 10/20 = 1/2. So E = (q/A) / (2e0) = 1/(4*e0) = 1/(4*8.854*10^-12) = 3.542*10^-11N/C.

 

[haruspex]

Thinking about it - I guess the 'charge density' must be the overall charge divided by the overall area - so 10/20 = 1/2. So E = (q/A) / (2e0) = 1/(4*e0) = 1/(4*8.854*10^-12) = 3.542*10^-11N/C.

No, that does the halving twice.
In your standard equation, the 2 in the denominator arises because the charge is considered equally split between the two sides, but the area only measures one side. You don't need to halve again.

 

[swinfen]

Ok - fair enough - so all of the information about the single point charge and its distance from the charged surface is moot (at least if we assume it is meant to be too small to matter for this level of course); the charge density is considered to be the charge over both surfaces (as stated), but only divided by the area of one surface (unless the question is just ambiguous and they mean the total surface area (i.e. of both sides) is 10m^2), then we can take that charge density and just use the formula: E = σ/2e0

 

[haruspex]

and they mean the total surface area (i.e. of both sides) is 10m^2)

No, you misunderstand me.
If the total charge on the plate is Q, and the plate has area A each side then the charge density each side is Q/(2A). You can think of the charge on each side as only sending out field lines on its own side, or as each side sending out half its field lines each side. It makes no difference to the field. Either way, you have a field $\frac Q{2A\epsilon_0}$ each side.
The standard formula you quote applies whether the plate is an insulator or a conductor, so doesn't discriminate the sides. $\sigma=Q/A$, $E=\frac{\sigma}{2\epsilon_0}=\frac Q{2A\epsilon_0}$