Question

What is the flux due to electric field E = 3*10^3icapN/C through a square of side 10cm, when it is held normal to E?

Answer 1

Explanation:

a)

Electric field intensity,

E

=3×10

3

N/C

Magnitude of electric field intensity, ∣E∣=3×10

3

N/C

Side of the square, s=10cm=0.1m

Area of the square, A=s

2

=0.01m

2

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ=0

o

Flux (ϕ) through the plane is given by the relation,

ϕ=∣

E

∣Acosθ

=3×10

3

×0.01×cos0

o

=30Nm

2

/C

(b)

Plane makes an angle of 60

o

with the x-axis. Hence, θ=60

o

Flux, ϕ=∣

E

∣Acosθ

=3×10

3

×0.01×cos60

o

=30×

2

1

=15Nm

2

/C

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