A charged cube

Electricity and Magnetism Level 5

Six identical square plates have surface charge density $\sigma=1 \mu C/m^2$ . What would be the force of interaction in Newtons between the plates if they are arranged to form a cube (the force acting on one face of the cube). The side of the plates is $L=10cm$ .

Details and assumptions

The vacuum permittivity is $\epsilon_{0}=8.85\times 10^{-12} F/m$ .

The answer is 0.000565.

3 solutions

Đinh Ngọc Hải
Sep 24, 2013

Because the cube is closed, the electric field outside near the surface is: $E_1=\frac{\sigma}{\epsilon_0}$ .

However, each plate itself induce a electric field near it: $E_2=\frac{\sigma}{2\epsilon_0}$ .

Therefore, the electric field induced by 5 other plates to one plate is: $E=E_1-E_2=\frac{\sigma}{2\epsilon_0}$ .

The force of interaction between the plates is: $F=Eq=\frac{\sigma}{2\epsilon_0} \sigma L^2=5.65*10^{-4} N$ .

If S is a cubical surface just larger than the cube of charged plates, with interior $V$ , then $\oint_S \mathbf{E}\cdot d\mathbf{S} \; = \; \int_V\tfrac{\rho}{\varepsilon_0}\,d\tau \; = \; \frac{6L^2\sigma}{\varepsilon_0}$ But the integral $F \;=\; \sigma \int \mathbf{E}\cdot d\mathbf{S}$ over any one of the six planes of the cube which comprise $S$ is equal to the normal component of the force (and, by symmetry, the whole force) on that charged plane due to the total electric field. This is the same for all six planes. Thus $F \; = \; \frac{L^2\sigma^2}{\varepsilon_0}$ Take off $E_2L^2\sigma = \tfrac{L^2\sigma^2}{2\varepsilon_0}$ , the contribution to this force from the portion of the electric field that is self-induced by the plate itself, and we get the result.

Mark Hennings - 7 years, 8 months ago

Can you please tell how we get electric field near the surface be $\frac{\sigma}{\in_{0}}$ .

I solved it using by first considering forces between small line charges and then solving lengthiest double integrals taking hours . Is there any standard integration type proof of that $\frac{\sigma}{\in_{0}}$ .

One could say it only if $E_{inside} = 0$ . How would we prove it.

jatin yadav - 7 years, 8 months ago

Well, I think it is classical. I read somewhere (can't remember exactly where) that the classic way is to construct an imaginary surface because we want to take advantage of the symmetry.

So let's orient a cylindrical Gaussian surface orthogonal to the surface. Remark first that the contribution to the electric flux is only from the top of the Gaussian surface. In fact then we can apply Gauss' Law to get that the flux is:

$ɸ$ = $EA = \frac{σA}{ε_0}$

And hence we derive the desired: $E$ = $\frac {σ}{ε_0}$

Anqi Li - 7 years, 8 months ago

please read the last line again, i meant i could also derive using gauss law but for that $E_{inside}$ must be $0$ .

jatin yadav - 7 years, 8 months ago

@Jatin Yadav – Hi are you asking why Ein is 0? Maybe you could consider constructing an imaginary surface just smaller than the cube. Because inside this volume, there is no charge, according to EA = q/ε0, Ein is 0.

Siyu Ren - 7 years, 8 months ago

Here's another proof. Because of the surface charge distribution, the normal component of any electric field must have a discontinuity of $\tfrac{\sigma}{\varepsilon_0}$ across one of the plates. Pick one plate $P$ . If $E_1$ is the field due to the other five plates, and $E_2$ is the field due to the plate itself, so that $E=E_1+E_2$ is the total field, then:

$E$ is zero inside the cube, and hence must have normal component $\tfrac{\sigma}{\varepsilon_0}$ near $P$ just outside the cube.
By symmetry, $E_2$ must have normal component $\tfrac{\sigma}{2\varepsilon_0}$ near $P$ just outside the cube, and $-\tfrac{\sigma}{2\varepsilon_0}$ near $P$ just inside the cube.

Subtracting these results, the field $E_1$ must have normal component $\tfrac{\sigma}{2\varepsilon_0}$ near $P$ both inside and outside the cube. Multiply this by $\sigma L^2$ for the answer.

Mark Hennings - 7 years, 8 months ago

Why is electric field zero inside the box?

Harsh Shrivastava - 3 years, 8 months ago

@Harsh Shrivastava – There is no electric field inside a sealed conducting box.

Mark Hennings - 3 years, 8 months ago

@Mark Hennings – Well it's not mentioned that box is conducting...

Harsh Shrivastava - 3 years, 8 months ago

@Harsh Shrivastava – How else can you have a charge distribution?

Mark Hennings - 3 years, 8 months ago

Rohan Singh
Sep 27, 2013

we take a pill box(very small cylindrical) Gaussian surface so that it just encloses a face. the area of the base is dA. the flux due to the curved surface is zero(as field is perpendicular to face there). now, we divide the box in 2 parts-the dA enclosed in Gaussian suface and the rest. therefore, flux through outer base is = (E(rest) + E(element))dA [this is as E outside is in same sense due to both] similarly, the sense of E(element)reverses inside and the flux is= (E(rest) - E(element))dA so equating net flux with σdA/ϵ , we get E(rest)=σ/2ϵ therefore dF=E(rest on element)(σdA) therefore F=(σ^2)(L^2)/2ϵ

i could not write a solution in the previous question so i just want to share it. in the previous question too, we can similarly comment on magnetic field at a point just outside and inside the cylinder, get expression of net magnetic field on the current carrying element and then simply write the force on the element wire due to the field. then by dividing by the area of the element wire, we get the pressure.

Ferris Prima Nugraha
Oct 11, 2017

The total charge of 6 sides of square is $q_{total} = 6 \sigma L^2$

Therefore, the electric field flux through each sides to be (remember to divide by 6):

$\phi = \frac{1}{6} \frac{q_{total}}{\epsilon_0} = \frac{\sigma^2 L^2}{\epsilon_0}$

Or by eliminating self-contribution of the charge, we have to divide it by 2

$\phi_{ext} = \frac{1}{2} \frac{\sigma^2 L^2}{\epsilon_0}$

The last thing is that solver must realise that the force exerted per side to be $F = \sigma \phi_{self}$

Which will give $F = \frac{\sigma^2 L^2}{2\epsilon_0} = 5.65 × 10^{-4} N$

A charged cube

The answer is 0.000565.

3 solutions

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