AIIMS 2005 Physics Electric Field MCQ Question

To solve the problem of the electric field between two infinitely long parallel conducting plates with surface charge densities +σ and -σ, we start by applying some fundamental concepts of electrostatics.

Explanation of the Correct Answer

1. Electric Field Due to a Single Plate: The electric field generated by an infinite plane sheet of charge with surface charge density $\sigma$ is given by the formula: $E = \frac{\sigma}{2\varepsilon_0}$ This field is directed away from the positively charged plate and towards the negatively charged plate.

2. Electric Fields from Both Plates:

   • For the positively charged plate (+σ), the electric field $\vec{E}_1$ in the space between the plates (toward the negative plate) is: $\vec{E}_1 = \frac{\sigma}{2\varepsilon_0} \text{ (directed to the right)}$
   • For the negatively charged plate (-σ), the electric field $\vec{E}_2$ in the space between the plates (also toward the negative plate) is: $\vec{E}_2 = \frac{\sigma}{2\varepsilon_0} \text{ (also directed to the right)}$

3. Net Electric Field Between the Plates: The total electric field $\vec{E}_{\text{total}}$ in the region between the plates is the vector sum of the electric fields due to both plates. Since both fields are in the same direction (toward the negative plate), we add their magnitudes: $\vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{2\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$

Thus, the electric field in the region between the plates is: $E = \frac{\sigma}{\varepsilon_0}$

This aligns with option C, which is the correct answer.

Clarification of Incorrect Options

• Option A: 0 volt/meter: This option implies that there is no electric field between the plates. However, we have established that there is a non-zero electric field due to the charges on the plates.

• Option B: $\frac{\sigma}{2\varepsilon_0}$ volt/meter: This value represents the electric field produced by a single plate only. Since we have two plates, we cannot simply take this value; we must consider the contributions from both plates, which adds up to $\frac{\sigma}{\varepsilon_0}$.

• Option D: $\frac{2\sigma}{\varepsilon_0}$ volt/meter: This value incorrectly suggests that both fields are added without considering their effective contributions in the direction of the field. As derived, the correct addition results in $\frac{\sigma}{\varepsilon_0}$, not doubling the field.

Summary

The electric field between two parallel conducting plates with surface charge densities +σ and -σ is determined by summing the contributions from both plates. This results in: $E = \frac{\sigma}{\varepsilon_0}$

This confirms option C as the correct answer. Understanding the contributions from both plates and the direction of the electric fields is crucial to solving such problems in electrostatics.