Physics Solution – Question 18

Solution to Question 18

1. Understanding the Forces on a Surface Charge

We have an insulating hollow cube of side $l$ with a uniform surface charge density $\sigma$. We need to find the force on one face (Face 1) due to the electric field created by the other 5 faces.

The total electric field $\vec{E}_{total}$ at any point on the surface of the face is the vector sum of:

$\vec{E}_{self}$: The field produced by the charge on Face 1 itself.
$\vec{E}_{others}$: The field produced by the remaining 5 faces.

Figure 1: Illustration of the force acting on an element of the charged face.

2. Field Boundary Conditions

Using the boundary conditions for the electric field near a charged surface with density $\sigma$: $$ E_{out} – E_{in} = \frac{\sigma}{\epsilon_0} $$ We can express the total field just outside ($E_{out}$) and just inside ($E_{in}$) in terms of the field from the local face ($E_{self}$) and the field from the rest of the cube ($E_{others}$).

Since $E_{self} = \frac{\sigma}{2\epsilon_0}$ (directed away from the face), we have: $$ E_{out} = E_{others} + \frac{\sigma}{2\epsilon_0} $$ $$ E_{in} = E_{others} – \frac{\sigma}{2\epsilon_0} $$ Solving for $E_{others}$, we get the average field: $$ E_{others} = \frac{E_{in} + E_{out}}{2} $$

3. Calculating the Force

The force $dF$ on a charge element $dq = \sigma dA$ is due only to $E_{others}$.

$$ dF = dq \cdot E_{others} = (\sigma dA) \left( \frac{\sigma}{2\epsilon_0} \right) = \frac{\sigma^2}{2\epsilon_0} dA $$

Integrating over the face area $A = l^2$:

$$ F = \int \frac{\sigma^2}{2\epsilon_0} dA = \frac{\sigma^2 l^2}{2\epsilon_0} $$

Aliter (Alternative Method)

Using Flux Balance on a Tight-Fitting Gaussian Surface

Consider a Gaussian surface that fits “tightly” around the cube in question.

1. Total Flux:
For the cube, the total charge enclosed is 6$q = 6\sigma A$. By Gauss’s Law, the total flux leaving one surface is: $$ \phi_{total} = \frac{q}{\epsilon_0} = \frac{\sigma A}{\epsilon_0} $$ This implies the total electric field just outside the surface is $E_{total} = \frac{\sigma}{\epsilon_0}$.

2. Separation of Fields:
This total field is the superposition of the field due to the face itself ($E_{face}$) and the field due to the remaining cube ($E_{cube}$). $$ E_{total} = E_{face} + E_{cube} $$

3. Contribution of the Face:
We know that for a plane sheet of charge, the field contribution due to the sheet itself is: $$ E_{face} = \frac{\sigma}{2\epsilon_0} $$

4. Contribution of the Remaining Cube:
Substituting the known values into the superposition equation: $$ \frac{\sigma}{\epsilon_0} = \frac{\sigma}{2\epsilon_0} + E_{cube} $$ $$ E_{cube} = \frac{\sigma}{\epsilon_0} – \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{2\epsilon_0} $$

5. Force Calculation:
The force on the face is determined by the interaction of its charge ($q = \sigma A$) with the external field provided by the rest of the cube ($E_{cube}$). $$ F = q E_{cube} = (\sigma A) \left( \frac{\sigma}{2\epsilon_0} \right) $$ Substituting $A = l^2$: $$ F = \frac{\sigma^2 l^2}{2\epsilon_0} $$

The force acting on a face due to all the other faces is: $$ \frac{\sigma^2 l^2}{2\epsilon_0} $$