A thin infinite sheet charge and an infinite line charge of respective charge densities +σ and +λ are placed parallel at 5 m distance from each other. Points 'P' and 'Q' are at 3/π m and 4/π m perpendicular distances from line charge towards sheet charge, respectively. 'E_P' and 'E_Q' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If Eₚ/E₀=4/a for 2|σ|=|λ|, then the value of a is ___________.

Solution

Physics sir · 4 steps

Between the line and the sheet the two fields point opposite ways, so subtract them, do not add.

1. Directions first. Both charges are positive. At a point between them, the line pushes the field toward the sheet and the sheet pushes it back toward the line. So the resultant is the difference of the two magnitudes. (The E₀ in the question means E_Q.)

   E = | λ/(2πε₀ r) - σ/2ε₀ |

2. Use λ = 2σ. The line field becomes 2σ/(2πε₀ r) = σ/(πε₀ r). The sheet field stays σ/2ε₀ at every distance.
3. At P and Q. At P, π r = 3, so E_P = σ/ε₀(1/2 - 1/3) = σ/6ε₀. At Q, π r = 4, so E_Q = σ/ε₀(1/2 - 1/4) = σ/4ε₀.
4. So. E_P/E_Q = (1/6)/(1/4) = 4/6. Comparing with 4/a gives a = 6.

The common slip: 3.6. Adding the two fields instead of subtracting gives E_P/E_Q = (5/6)/(3/4) = 10/9, and so a = 3.6 instead of 6.

The idea to keep. Before you add fields, mark each one's direction at the point. Two like charges point opposite ways between them. A sheet's field does not depend on distance. A line's field falls as 1/r.

Answer from the NTA final answer key. Solution by PrepWiser.