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Question
Obtain an expression for electric field intensity at a point outside uniformly charged thin plane sheet.
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Solution
Expression for electric intensity due to uniformly charged infinite plane sheet:
a. Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σ on both sides of the sheet.
b. By symmetry, it follows that the electric field is perpendicular to the plane sheet of charge and is directed in outward direction.
c. Electric field intensity has same magnitude at a given distance on either sides of the sheet.
d. To find electric field intensity at a point P due to uniformly charged infinite thin plane sheet, construct an imaginary cylinder around P with its axis perpendicular to plane sheet carrying charge with ends having cross sectional area ds.
e. The plane sheet passes through the middle of cylinder’s length so that the ends of cylinder are equidistant from the plane sheet carrying charge.
f. Electric field intensity, `vecE` is perpendicular to the ends of cylinder, hence the electric flux through each end is Eds.
g. Since `vecE` is perpendicular to plane sheet, it is parallel to the curved surface of Gaussian cylinder. Hence, electric flux does not pass through the curved surface of Gaussian cylinder.
h. Now, Total Normal Electric Induction over Gaussian surface = ε E (2ds)
where, ds is surface area of end faces of the cylinder.
Algebraic sum of charges enclosed by Gaussian cylinder = σds
According to Gauss’ law,
ε E (2ds) = σds
E =σ/2ε
This is the expression for electric field intensity at a point outside uniformly charged thin plane sheet.
j. Above equation shows that the magnitude of electric field intensity is independent of the distance of point from plane sheet.
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