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lutik1710 [3]
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?
Physics
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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