Please,give solution

Asked by archanajain7219 | 6th May, 2022, 10:45: 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/ (4πεo ) 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 εo )

Answered by Thiyagarajan K | 8th May, 2022, 12:32: AM

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