Obtain an expression for electric field due to an infinite charged plane sheet
Answers
Answer:
E=σ2ϵ0. This is independent of the distance of P from the infinite charged sheet. The electric field lines are uniform parallel lines extending to infinity.
Answer:
E = sigma/(2 epsilon0)
Explanation:
please see the enclosed attachment for detailed explanation and step by step working.
Let us treat the infinitely charged sheet as a combination of infinite number of concentric rings. of radius r and thickness dr, Let's say the area charge density be sigma Coul/m^2.
K = 1/(4π epsilon_0)
Electric field dE1 due to a small area dA
= r d phi × dr is = K sigma r d phi dr /(d^2 Sec^2 theta)
Electric field dE due to dE1 & dE2 along the axis of the ring
= 2 K sigma r dr Cos^2 theta /d^2
Electric field dE due to the thin ring of radius r and thickness dr at a point P which is at a distance d from the sheet
= 2πK sigma r dr Cos^2 theta /d^2
now r = d tan theta. dr = d sec^2 theta d theta.
dE = [sigma/(2 epsilon_0)] × Sin theta d theta.
Integrating dE over r =0 to infty, theta from 0 to π/2,
E = sigma/(2 epsilon_0)
It's independent of the distance d between the charged sheet and the point P.
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Using Gauss theorem derive an expression for electric field due to a uniformly charge infinite plane sheet.?
please see the enclosed attachment for detailed explanation and step by step working.
take an infinitely large charged sheet MNOP of charge density sigma.
Take a point P at the centre of EFGH. To find E at P, construct a Gaussian surface - a cuboid IJKL - EFGH . The electric field and flux lines flow out of the sheet perpendicular to it.
Gauuss's law:
closed Integral over cuboid of E • dA = q/epsilon0.
2 E = sigma/ epsilon0
E = sigma/(2 epsilon0
