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Asked by archanajain7219 | 06 May, 2022, 22:45: PM

Expert Answer

The question understood is , a charge q is placed at one side of semi-infinite plate .

The charge q is at a distance d from the bottom of the plate.

Electric flux passing through the plate is to be determined.

Let us consider a small area dA at a distance r from bottom edge of plate .

Electric field E due to charge q at this area is given as

begin mathsize 14px style E space equals space K cross times fraction numerator q over denominator left parenthesis space d squared plus r squared space right parenthesis end fraction end style

where K = 1/ (4πε_o ) is Coulomb's constant .

Flux dφ through area is given as , dφ = E dA cosθ

where θ is the angle between electric field direction and norma to the area as shown in figure

begin mathsize 14px style F l u x space d phi space equals space K space cross times space fraction numerator q over denominator open parentheses d squared plus r squared close parentheses end fraction cross times space d A space cross times space cos theta end style

begin mathsize 14px style F l u x space d phi space equals space K space cross times space fraction numerator q cross times d over denominator open parentheses d squared plus r squared close parentheses to the power of 3 divided by 2 end exponent end fraction cross times space d A space end style

Entire flux passing through plate is determined by integrating above expression in r-θ coordinate system

begin mathsize 14px style phi space equals space integral integral d phi space equals space K cross times q cross times d cross times integral subscript 0 superscript pi d theta space integral subscript 0 superscript infinity fraction numerator r space d r over denominator open parentheses d squared plus r squared close parentheses to the power of 3 divided by 2 end exponent end fraction end style

by performing above integration, we get

φ = K × q × π = q / ( 4 ε_o )

Answered by Thiyagarajan K | 08 May, 2022, 00:32: AM

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