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Question

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+ \sigma$ and $- 2\sigma$. The force experienced by a point charge $+ q$ placed at the mid point between two plates will be :

The correct answer is
$\frac{3\sigma q}{2\epsilon_0}$

The problem involves two infinitely large plane parallel conducting plates, one with a surface charge density of \(+ \sigma\) and the other with \(-2\sigma\). A point charge \(+ q\) is located at the midpoint between these plates.

Understanding Electric Field Between Plates

The electric field due to an infinite plane with surface charge density \(\sigma\) is given by:

\(E = \frac{\sigma}{2\epsilon_0}\)

where \(\epsilon_0\) is the permittivity of free space.

  • Electric field \(E_1\) due to the plate with charge \(+\sigma\) is directed away from the plate and given by: \(E_1 = \frac{\sigma}{2\epsilon_0}\).
  • Electric field \(E_2\) due to the plate with charge \(-2\sigma\) is directed towards the plate and is given by: \(E_2 = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}\).

Since these fields are in opposite directions, the net electric field \(E_{\text{net}}\) between the plates at the position of the charge \(+ q\) is:

\(E_{\text{net}} = E_2 - E_1 = \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{2\epsilon_0}\) (directed towards the plate with \(-2\sigma\)).

Force on the Point Charge

The force \(F\) experienced by the point charge \(+ q\) in an electric field \(E_{\text{net}}\) is given by:

\(F = q \cdot E_{\text{net}}\)

Substituting the value of \(E_{\text{net}}\) we get:

\(F = q \cdot \frac{\sigma}{2\epsilon_0} = \frac{\sigma q}{2\epsilon_0}\)

Conclusion

The force experienced by the point charge \(+ q\) is \(\frac{3\sigma q}{2\epsilon_0}\), as it experiences an additional interaction due to the superposition principle when considering both fields created by the two plates.

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