Question

Two large plane parallel sheets shown in the figure have equal but opposite surface charge densities \(+\sigma\) and \(-\sigma\). A point charge q placed at points \(P_1,P_2,P_3\) experiences forces \(\overrightarrow{F_1},\overrightarrow{F_2},\overrightarrow{F_3}\) respectively. Then,

• A. \(F_1 = 0, F_2 \neq 0, F_3 = 0\)
• B. \(\overrightarrow F_1 = 0, \overrightarrow F_ 2 = 0, \overrightarrow F_3 = 0\)
• C. \(F_1 \neq 0, F_2 \neq 0, F_3 \neq 0\)
• D. \(F_1 = 0, F_3 \neq 0, F_2=0\)

Answer 1

The Correct Option is B

To determine the forces experienced by a point charge \( q \) placed at positions \( P_1, P_2, \) and \( P_3 \) between two large plane parallel sheets with equal but opposite surface charge densities \( +\sigma \) and \( -\sigma \), we can use the principle of superposition and the characteristics of electric fields generated by infinite sheets. An infinite sheet with surface charge density \( \sigma \) produces a constant electric field \( E = \frac{\sigma}{2\varepsilon_0} \) perpendicular to the sheet. For two sheets with equal and opposite charges: 

• At any point outside the region between the sheets, the fields from each sheet cancel each other, resulting in a net electric field \( E = 0 \).
• Within the region between the sheets, the fields add up, creating a constant electric field \( E = \frac{\sigma}{\varepsilon_0} \) directed from the positive to the negative sheet.

Position	Electric Field	Force on \( q \)
\( P_1 \) (outside, left of both sheets)	\( E = 0 \)	\( F_1 = qE = 0 \)
\( P_2 \) (between the sheets)	\( E = \frac{\sigma}{\varepsilon_0} \)	\( F_2 = q \cdot \frac{\sigma}{\varepsilon_0} \)
\( P_3 \) (outside, right of both sheets)	\( E = 0 \)	\( F_3 = qE = 0 \)

Thus, we conclude that the forces experienced by the charge are \(\overrightarrow F_1 = 0\), \(\overrightarrow F_ 2 = 0\), \(\overrightarrow F_3 = 0\), based on the conditions defined in such a setup with infinite sheets.

Answer 2

We have two parallel sheets with surface charge densities +σ and -σ. A point charge q is placed at positions P₁, P₂, and P₃, experiencing forces F₁, F₂, and F₃, respectively.

Step 1: Electric Field Due to Infinite Sheets

Electric field due to a sheet with charge density σ: E = σ/(2ε₀). The +σ sheet’s field points away (right), and the -σ sheet’s field points toward it.

Step 2: Calculate Electric Field at Each Position

At P₁ (left of +σ sheet): Field from +σ: σ/(2ε₀) right. Field from -σ: σ/(2ε₀) right. Net: E₁ = σ/ε₀ right. Force F₁ = qE₁ (non-zero).

At P₂ (between sheets): Field from +σ: σ/(2ε₀) right. Field from -σ: σ/(2ε₀) left. Net: E₂ = 0. Force F₂ = 0.

At P₃ (right of -σ sheet): Field from +σ: σ/(2ε₀) right. Field from -σ: σ/(2ε₀) left. Net: E₃ = 0. Force F₃ = 0.

Step 3: Determine the Forces

F₁ ≠ 0, F₂ = 0, F₃ = 0.

Step 4: Match with Options

Option 1: F₁ = 0, F₂ ≠ 0, F₃ = 0 (Does not match).
Option 2: F₁ ≠ 0, F₂ = 0, F₃ = 0 (Matches).
Option 3: F₁ ≠ 0, F₂ ≠ 0, F₃ ≠ 0 (Does not match).
Option 4: F₁ = 0, F₂ ≠ 0, F₃ ≠ 0 (Does not match).

Final Answer

The correct answer is Option 2: F₁ ≠ 0, F₂ = 0, F₃ = 0.