ElectrostaticsMedium

Question

Imagine a situation, in which an infinite sheet with positive charge +σ per unit area lies in the xy-plane and a second infinite sheet with negative charge −σ per unit area lies in the yz-plane. The net electric field E at any point (x, y, z) [that does not lie on either of these planes xy or yz] can be expressed as

Options

A.E = (σ/2ε0)(−î + k̂)
B.E = (σ/2ε0)ĵ
C.E = (σ/2ε0)[−(x/|x|)î + (z/|z|)k̂]
D.E = (σ/ε0)[(x/|x|)î − (z/|z|)k̂]

Solution

[Key Concept: An infinite sheet gives a uniform field σ/(2ε0) normal to it, away from positive and towards negative charge, reversing on crossing the sheet.]

Step 1: The +σ sheet (xy-plane, z = 0) gives (σ/2ε0)(z/|z|)k̂: along +k̂ above it and −k̂ below it.

Step 2: The −σ sheet (yz-plane, x = 0) points towards the sheet: −(σ/2ε0)(x/|x|)î.

Step 3: Superposing: E = (σ/2ε0)[−(x/|x|)î + (z/|z|)k̂]; a fixed vector such as Option A cannot be right in all regions.

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