Question

A surface has the area vector A=(2i+3j) m2. What is the flux of a uniform electric field through the area if the field is (a) E = 4i^ N/C and (b) E = 4kˆ N/C?​

Answer 1

Answer:

Flux ==E. A ie for (a) 4î.(2î+3j)=8+0

=8 N.M²/C,(b)4k.(2î+3j)=0.0=0.Ans..

Answer 2

a) Φ = -8 m^2/C + 12k^ m^2/C

b) 8k^ m^2/C

The flux of a uniform electric field through a surface is given by the dot product of the electric field and the surface area vector:

Φ = E · A

where Φ is the flux, E is the electric field, and A is the surface area vector.

a) If the electric field is E = 4i^ N/C, then the flux through the surface is:

Φ = E · A = (4i^ N/C) · (2i+3j m^2) = 8i^2 m^2/C + 12ij m^2/C

Note that i^2 = -1, so we can simplify the first term:

Φ = -8 m^2/C + 12ij m^2/C

The second term involves the cross product of the unit vectors i and j, which is given by k^, so we can simplify further:

Φ = -8 m^2/C + 12k^ m^2/C

b) If the electric field is E = 4k^ N/C, then the flux through the surface is:

Φ = E · A = (4k^ N/C) · (2i+3j m^2) = 8ik^ m^2/C + 12jk^ m^2/C

Note that i·k^ = j·k^ = 0, so we can simplify the expression:

Φ = 8ik^ m^2/C + 12jk^ m^2/C = 8k^ m^2/C

Therefore, the flux through the surface is 8 m^2/C if the electric field is in the z-direction (i.e., perpendicular to the surface).

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