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3 years ago

8

A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag

nitude of the electric flux through the surface is 78 N · m2/C. What is the angle (less than 90) between the direction of the electric field and thenormal to the surface?

1 answer:

klio [65]3 years ago

4 0

Answer:

57.94°

Explanation:

we know that the expression of flux

$\Phi =E\times S\times COS\Theta$

where Ф= flux

E= electric field

S= surface area

θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 $\frac{Nm^{2}}{sec}$

E= $1.44\times 10^{4}\frac{Nm}{C}$

S= $\pi \times 0.057^{2}$

$COS\Theta =\frac{\Phi }{S\times E}$

= $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

=0.5306

θ=57.94°

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