JEE Main 2025 Apr 2 Shift 1, Physics Q5: Coulomb's Law And Electric Field

JEE Main2025Apr 2, Shift 1Physics

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+ σ$ and $- 2 σ$ . The force experienced by a point charge $+ q$ placed at the mid point between two plates will be :

A
$\frac{σ q}{4 ϵ _{0}}$
B
$\frac{3 σ q}{2 ϵ _{0}}$
C
$\frac{3 σ q}{4 ϵ _{0}}$
D
$\frac{σ q}{2 ϵ _{0}}$

Solution

For two parallel conducting plates with given net surface charge densities, the charges redistribute on the four faces so that the field inside each conductor is zero.

Standard redistribution result: for plates with total surface densities $σ_{1}$ (left) and $σ_{2}$ (right), the four faces carry, from left to right:

$\frac{σ _{1} + σ _{2}}{2}, \frac{σ _{1} - σ _{2}}{2}, \frac{σ _{2} - σ _{1}}{2}, \frac{σ _{1} + σ _{2}}{2}$

With $σ_{1} = + σ$ and $σ_{2} = - 2 σ$ , the four faces are:

$- \frac{σ}{2}, \frac{3 σ}{2}, - \frac{3 σ}{2}, - \frac{σ}{2}$

Field at the midpoint (between the inner faces) comes from these inner faces, which together act as two parallel infinite sheets. The net field magnitude between them:

$E = \frac{σ _{inner,left} - σ _{inner,right}}{2 ϵ _{0}} = \frac{\frac{3 σ}{2} - ( - \frac{3 σ}{2} )}{2 ϵ _{0}} = \frac{3 σ}{2 ϵ _{0}}$

Force on the test charge:

$F = q E = \frac{3 σ q}{2 ϵ _{0}}$

$∴$ Option (2) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q5: Coulomb's Law And Electric Field

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