Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point P is (xσ)/(ε₀). The value of x is _________ (all quantities are measured in SI units). Numerical answer type. Work it out, then open the answer.

Answer: 2. Each sheet gives (|σ|)/(2ε₀): away from a positive sheet, towards a negative sheet. P lies between x = a and x = 3a. Sheet -σ at x = -a (left of P): towards it, -x, magnitude (σ)/(2ε₀). Sheet -2σ at x = a (left of P): towards it, -x, magnitude (σ)/(ε₀). Sheet +σ at x = 3a (right of P): away from it, -x, magnitude (σ)/(2ε₀). All along -x: E = (σ)/(2ε₀) + (σ)/(ε₀) + (σ)/(2ε₀) = (2σ)/(ε₀) So x = 2.

Three infinitely long charged thin sheets are placed as shown in… | JEE Main 2024 (6 Apr, Shift 1)

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Electric Flux & Gauss's Law

Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point $P$ is $\frac{x\sigma}{\epsilon_0}$ . The value of $x$ is _________ (all quantities are measured in SI units).

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Each sheet gives $\frac{|\sigma|}{2\varepsilon_0}$ : away from a positive sheet, towards a negative sheet. P lies between $x = a$ and $x = 3a$ .
Sheet $-\sigma$ at $x = -a$ (left of P): towards it, $-x$ , magnitude $\frac{\sigma}{2\varepsilon_0}$ .
Sheet $-2\sigma$ at $x = a$ (left of P): towards it, $-x$ , magnitude $\frac{\sigma}{\varepsilon_0}$ .
Sheet $+\sigma$ at $x = 3a$ (right of P): away from it, $-x$ , magnitude $\frac{\sigma}{2\varepsilon_0}$ .
All along $-x$ : $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{2\sigma}{\varepsilon_0}$
So $x = 2$ .

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