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Questions  >>  CBSE XII  >>  Math  >>  Model Papers
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Consider a uniform electric field $E=3 \times 10 \times 10^3 \hat i N/C$. What is the flux of this foeld through a square of $10 \;cm$ a side whose plane is parallel to the y-z plane ?

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A)
Answer :
Given electric field $E= 3 \times 10^3 \hat i N/C$
(ie) electric field is directed towards x-axis
As the surface is in the y-z plane the are vector is along the x- axis .
area vector is the normal to the square .
Area= $10 \times 10 =100 \;.sq.cm$
$\qquad= 10^{-2}\;m^2$
Area vector $S= 10^{-2} C m^2$
Using the formula
$\phi = E. S =ES \cos \theta$
$\qquad= E.S$
Because angle between E and S is $0^{\circ}$
$\cos 0 =1$
$\therefore \phi = 3 \times 10^3 \times 10^{-2}$
$\qquad= 30 \;N-m^2/c$
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