Question

Consider a uniform electric field E = 3 × 103 î N/C. (a) What is theflux of this field through a square of 10 cm on a side whose plane isparallel to the yz plane? (b) What is the flux through the samesquare if the normal to its plane makes a 60° angle with the x-axis?

Answer 1

Given: The side of the square is 10 cm and uniform electric field is 3× 10 3 i ^   NC −1 .

The area of the square is given as,

A= s 2

Where, s is the side of the square.

By substituting the values in the above expression, we get

A= 0.1 2 =0.01  m 2

The electric flux through a plane is given as,

ϕ=| E |Acosθ (1)

Where, E is the electric field and θ is the angle between the electric field and the unit vector normal to the plane.

a)

Since, the plane of the square is parallel to y-z plane, thus angle between unit vector normal to the plane and electric field will be 0°.

By substituting the values in the equation (1), we get

ϕ=3× 10 3 ×0.01×cos0° =30  Nm 2 C −1

Thus, electric flux through the plane is 30  Nm 2 C −1 .

b)

Since, the plane makes 60° with the x-axis, thus θ=60°.

By substituting the values in the equation (1), we get

ϕ=3× 10 3 ×0.01×cos60° =15  Nm 2 C −1

Thus, the electric flux through the plane will be 15  Nm 2 C −1 .