Question

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

• A. \( \frac{3q\sigma}{4 \epsilon_0} \)
• B. \( \frac{3q\sigma}{2 \epsilon_0} \)
• C. \( \frac{3q\sigma}{4 \epsilon_0} \)
• D. \( \frac{q\sigma}{2 \epsilon_0} \)

Hint:

When calculating the electric field due to uniformly charged infinite plates, use the formula \( E = \frac{\sigma}{2\epsilon_0} \) for each plate and then sum the fields considering the direction.

Answer 1

The Correct Option is B

Let the charge distribution on the two plates be \( \sigma \) and \( -\sigma \), with the point charge \( q \) placed at the midpoint between the plates.
The electric field due to each plate at the midpoint is as follows:  
For Plate 1, the electric field is \( \frac{\sigma}{2 \epsilon_0} \) directed away from the plate, and for Plate 2, the electric field is \( \frac{\sigma}{2 \epsilon_0} \) directed towards the plate. 
Thus, the net electric field experienced by the charge \( q \) is: \[ E_{\text{net}} = \frac{3\sigma}{2 \epsilon_0} \] Now, the force on the charge \( q \) is given by: \[ F = qE = q \times \frac{3\sigma}{2 \epsilon_0} = \frac{3q\sigma}{2 \epsilon_0} \]

Answer 2

The problem asks for the force experienced by a point charge \(+q\) placed at the midpoint between two infinitely large, parallel conducting plates. The plates have uniform surface charge densities of \(+\sigma\) and \(-2\sigma\) respectively.

Concept Used:

The solution utilizes the principle of superposition for electric fields and the formula for the electric field due to an infinite conducting sheet of charge.

1. Electric Field of an Infinite Conducting Plate: The electric field \( E \) at any point due to an infinite conducting plate with surface charge density \( \sigma_{plate} \) is uniform (independent of distance from the plate) and has a magnitude of: \[ E = \frac{|\sigma_{plate}|}{2\epsilon_0} \] where \( \epsilon_0 \) is the permittivity of free space. The direction of the field is perpendicular to the plate; it points away from a positively charged plate and towards a negatively charged plate.
2. Superposition Principle: The net electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge at that point.
3. Electric Force: A charge \( q \) placed in an electric field \( \vec{E} \) experiences a force \( \vec{F} \) given by: \[ \vec{F} = q\vec{E} \]

Step-by-Step Solution:

Step 1: Determine the electric field at the midpoint due to the plate with charge density \(+\sigma\).

Let's denote the left plate with charge density \( \sigma_1 = +\sigma \) as Plate 1. Since it is positively charged, the electric field \( \vec{E}_1 \) it produces will be directed away from it, i.e., to the right.

The magnitude of this electric field is:

\[ E_1 = \frac{\sigma_1}{2\epsilon_0} = \frac{\sigma}{2\epsilon_0} \]

Step 2: Determine the electric field at the midpoint due to the plate with charge density \(-2\sigma\).

Let's denote the right plate with charge density \( \sigma_2 = -2\sigma \) as Plate 2. Since it is negatively charged, the electric field \( \vec{E}_2 \) it produces will be directed towards it, i.e., also to the right.

The magnitude of this electric field is:

\[ E_2 = \frac{|\sigma_2|}{2\epsilon_0} = \frac{|-2\sigma|}{2\epsilon_0} = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \]

Step 3: Calculate the net electric field at the midpoint.

Since both electric fields \( \vec{E}_1 \) and \( \vec{E}_2 \) point in the same direction (to the right), the net electric field \( \vec{E}_{net} \) is the sum of their magnitudes.

\[ E_{net} = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{\epsilon_0} \] \[ E_{net} = \frac{\sigma + 2\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0} \]

The direction of the net electric field is to the right, from Plate 1 towards Plate 2.

Final Computation & Result:

Step 4: Calculate the force experienced by the point charge \(+q\).

The force \( \vec{F} \) on the charge \(+q\) is given by the formula \( \vec{F} = q\vec{E}_{net} \).

\[ F = (+q) \cdot E_{net} = q \left( \frac{3\sigma}{2\epsilon_0} \right) = \frac{3q\sigma}{2\epsilon_0} \]

Since the charge \( q \) is positive, the direction of the force is the same as the direction of the net electric field, which is towards the right.

Therefore, the force experienced by the point charge is \( \frac{3q\sigma}{2\epsilon_0} \) directed towards the plate with charge density \( -2\sigma \).