Aflat surface of area 3.20 m2 is rotated in a uniform electric field of magnitude e = 6.20 x 105 n/c. determine the electric flux through this area (a) when the electric field is perpendicular to the surface and (b) when the electric field is parallel to the surface.

The electric flux through this area:

This is defined as the number of electric field lines that intersect a given area.

Electric flux(φ)= EA cos θ

where θ = angle between the directions of E and A is the surface area.

When the electric field is perpendicular to the surface,  θ = 0°.

φ = EA cos(0°)  = (6.20 x 10⁵ N/C)*(3.2 m²) × 1 = 1.984 x 10⁶ Nm²/C

When the electric field is parallel to the area, then θ = 90°, and

φ = EA cos(90°) = 0 as result of cos 90° being zero.

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double the distance = divide strength by 4

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.