1. Electric field due to an infinite charged conducting plate

For a conductor, the electric field just outside its surface is $E = \frac{\sigma_{\text{surface}}}{\epsilon_0}$ where $\sigma_{\text{surface}}$ is the charge density on that face.

Also, inside a conductor, the electric field is zero.

2. Let the surface charge densities on the inner faces be

• Left plate inner face: $\sigma_1$
• Right plate inner face: $\sigma_2$

Given total charge densities on the plates are:

• Left plate: $+\sigma$
• Right plate: $-2\sigma$

Since the outer faces may also carry charge, let outer face densities be:

• Left outer face: $\sigma_3$
• Right outer face: $\sigma_4$

Then total charge on each plate gives: $\sigma_1 + \sigma_3 = +\sigma$ $\sigma_2 + \sigma_4 = -2\sigma$

3. Use the condition that field inside each conductor is zero

For two infinite conducting plates, the charges on the facing surfaces must be equal and opposite: $\sigma_1 = -\sigma_2$

Now, for isolated plates, the remaining charge distributes on the outer surfaces. Solving with total charges:

Let $\sigma_1 = x, \quad \sigma_2 = -x$ Then $\sigma_3 = \sigma - x$ $\sigma_4 = -2\sigma + x$

For infinite sheet configuration, the electric field far outside on the left depends on total charge: $E_{\text{left outside}} = \frac{\sigma + (-2\sigma)}{2\epsilon_0} = -\frac{\sigma}{2\epsilon_0}$ And just outside the left outer surface of conductor, $E = -\frac{\sigma_3}{\epsilon_0}$ so $\sigma_3 = \frac{\sigma}{2}$ Hence $x = \sigma - \frac{\sigma}{2} = \frac{\sigma}{2}$ Thus $\sigma_1 = \frac{\sigma}{2}, \qquad \sigma_2 = -\frac{\sigma}{2}$

So the charges on the inner faces are $+\frac{\sigma}{2}$ and $-\frac{\sigma}{2}$.

4. Field in the region between the plates

Between the plates, both inner faces produce field in the same direction (from positive to negative plate).

Magnitude of field due to left inner face: $E_1 = \frac{\sigma/2}{\epsilon_0} = \frac{\sigma}{2\epsilon_0}$

Magnitude of field due to right inner face: $E_2 = \frac{\sigma/2}{\epsilon_0} = \frac{\sigma}{2\epsilon_0}$

Therefore total field between plates is $E = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$

Direction: from the $+\sigma$ plate toward the $-2\sigma$ plate.

5. Force on charge $+q$ at the midpoint

$F = qE = q\cdot \frac{\sigma}{\epsilon_0} = \frac{\sigma q}{\epsilon_0}$

6. Compare with options

The derived force is $\boxed{\frac{\sigma q}{\epsilon_0}}$

This does not match any of the given options.

So the stored answer $\text{B} = \frac{3\sigma q}{2\epsilon_0}$ is not correct for conducting plates.

If the question had been about non-conducting infinite sheets with charge densities $+\sigma$ and $-2\sigma$, then field at midpoint would be $E = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$ so $F = \frac{3\sigma q}{2\epsilon_0}$ which matches option B.

Thus, there is likely an error in the wording: "conducting" should probably be "charged sheets" or "non-conducting sheets."