Question:
In the figure, a very large plane sheet of positive charge is shown. \(P_1\) and \(P_2\) are two points at distance \(l \) and \(2l\) from the charge distribution. If \(σ \) is the surface charge density, then the magnitude of electric fields \(E_1\) and \(E_2\) and \(P_1\) and \(P_2\) respectively are
In the figure, a very large plane sheet of positive charge is shown. \(P_1\) and \(P_2\) are two points at distance \(l \) and \(2l\) from the charge distribution. If \(σ \) is the surface charge density, then the magnitude of electric fields \(E_1\) and \(E_2\) and \(P_1\) and \(P_2\) respectively are
Updated On: Nov 21, 2025
- \(E_1 = \frac {σ}{ε_0}, E_2 = \frac {σ}{2ε_0}\)
- \(E_1 = \frac {2σ}{ε_0}, E_2 = \frac {σ}{ε_0}\)
- \(E_1 = E_2 = \frac {σ}{2ε_0}\)
- \(E_1 = E_2 = \frac {σ}{ε_0}\)
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The Correct Option is C
Solution and Explanation
For an infinite charged plane
\(E = \frac {σ}{2ε_0}\)
for each value of distance l
\(E_1 = \frac {σ}{2ε_0}\)
\(E_2 = \frac {σ}{2ε_0}\)
Thus,
\(E_1 = E_2 = \frac {σ}{2ε_0}\)
So, the correct option is (C): \(E_1 = E_2 = \frac {σ}{2ε_0}\)
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).