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Annette [7]
3 years ago
11

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

form electric field. What is the magnitude of the electric field if the field points 31.6° above the horizontal?
Physics
1 answer:
Tresset [83]3 years ago
4 0

Answer:231.16 N/C

Explanation:

Given

Electric Flux=147 N-m^2/C

Area(A)=0.824 m^2

Given Field point above 31.6 ^{\circ}

Therefore angle between Area vector Electric Field =90-31.6=58.4^{\circ}

We know that Flux is given by

\phi =\vec{E}\cdot \vec{A}

\phi =EAcos\theta

147=E\times 0.824\times cos(58.4)

E=231.16 N/C

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