Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ and -2σ. The force experienced by a point charge +q placed at the mid point between two plates will be: (a) (3σ q)/(4ε₀) (b) (3σ q)/(2ε₀) (c) (σ q)/(4ε₀) (d) (σ q)/(2ε₀)

Answer: (b). +σ−2σ+qσ/2ε₀σ/ε₀ Each plate acts as a charged sheet: E = (σ)/(2ε₀). +σ plate: (σ)/(2ε₀) away from it (towards the right). -2σ plate: (2σ)/(2ε₀) = (σ)/(ε₀) towards it (also to the right). E = (3σ)/(2ε₀), so F = qE = (3σ q)/(2ε₀), answer (b).

Consider two infinitely large plane parallel conducting plates as… | JEE Main 2025 (2 Apr, Shift 1)

Q28JEE Main 2025 · 2 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

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:

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Answer: (b)

Each plate acts as a charged sheet: $E = \frac{\sigma}{2\varepsilon_0}$ .
$+\sigma$ plate: $\frac{\sigma}{2\varepsilon_0}$ away from it (towards the right).
$-2\sigma$ plate: $\frac{2\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$ towards it (also to the right).
$E = \frac{3\sigma}{2\varepsilon_0}$ , so $F = qE = \frac{3\sigma q}{2\varepsilon_0}$ , answer (b).

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Electric Flux & Gauss's Law question from JEE Main 2025 (2 Apr, Shift 1)

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