Question

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 \(\frac{3}{π} \)m and \(\frac{4}{π}\) m perpendicular distance 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 \(\frac{E_p}{E_Q} =\frac{ 4}{a}\) for 2|σ| = |λ|. Then the value of a is _______.

Hint:

When dealing with infinite sheet and line charges, use the respective electric field formulas and simplify using given conditions for charge densities and distances.

Answer 1

Correct Answer: 6

Step 1: Electric Field Due to Infinite Sheet
The electric field due to an infinite sheet charge is given by: \[ E_\text{sheet} = \frac{\sigma}{2\epsilon_0}. \] Step 2: Electric Field Due to Infinite Line Charge
The electric field due to an infinite line charge at a perpendicular distance \( r \) is given by: \[ E_\text{line} = \frac{\lambda}{2\pi \epsilon_0 r}. \] Step 3: Electric Fields at Points 'P' and 'Q'
For point 'P', the distance from the line charge is \( r_P = \frac{3}{\pi} \). The resultant electric field is: \[ E_P = E_\text{sheet} + E_\text{line} = \frac{\sigma}{2\epsilon_0} + \frac{\lambda}{2\pi \epsilon_0 r_P}. \] For point 'Q', the distance from the line charge is \( r_Q = \frac{4}{\pi} \). The resultant electric field is: \[ E_Q = E_\text{sheet} + E_\text{line} = \frac{\sigma}{2\epsilon_0} + \frac{\lambda}{2\pi \epsilon_0 r_Q}. \] Step 4: Ratio of Electric Fields
The ratio \( \frac{E_P}{E_Q} \) is given by: \[ \frac{E_P}{E_Q} = \frac{\frac{\sigma}{2\epsilon_0} + \frac{\lambda}{2\pi \epsilon_0 r_P}}{\frac{\sigma}{2\epsilon_0} + \frac{\lambda}{2\pi \epsilon_0 r_Q}}. \] Substitute \( 2|\sigma| = |\lambda| \), \( r_P = \frac{3}{\pi} \), and \( r_Q = \frac{4}{\pi} \): \[ \frac{E_P}{E_Q} = \frac{\frac{\lambda}{4\epsilon_0} + \frac{\lambda}{2\pi \epsilon_0 \cdot \frac{3}{\pi}}}{\frac{\lambda}{4\epsilon_0} + \frac{\lambda}{2\pi \epsilon_0 \cdot \frac{4}{\pi}}}. \] Step 5: Simplify the Expression
Simplify the numerator and denominator: \[ \frac{E_P}{E_Q} = \frac{\frac{\lambda}{4\epsilon_0} + \frac{\lambda}{6\epsilon_0}}{\frac{\lambda}{4\epsilon_0} + \frac{\lambda}{8\epsilon_0}}. \] Combine the terms: \[ \frac{E_P}{E_Q} = \frac{\frac{3\lambda}{12\epsilon_0} + \frac{2\lambda}{12\epsilon_0}}{\frac{2\lambda}{8\epsilon_0} + \frac{\lambda}{8\epsilon_0}} = \frac{\frac{5\lambda}{12\epsilon_0}}{\frac{3\lambda}{8\epsilon_0}}. \] Simplify further: \[ \frac{E_P}{E_Q} = \frac{5}{12} \times \frac{8}{3} = \frac{40}{36} = \frac{10}{9}. \] Equating \( \frac{10}{9} = \frac{4}{a} \), solve for \( a \): \[ a = 6. \]