Question

Expression for an electric field is given by $\overrightarrow{ E }=4000 x^2 i \frac{ V }{ m }$ The electric flux through the cube of side $20 cm$ when placed in electric field (as shown in the figure) is ___$V cm$

Hint:

The electric flux is the product of the electric field and the area through which the field lines pass, considering the angle between the electric field and the area vector.

Answer 1

Correct Answer: 640

The correct answer is 640
Flux \(=\vec{E}⋅\vec{A} \)
\(=4000(0⋅2)^2\frac{V}{m}​⋅(0⋅2)^2m^2 \)
=4000×16×10⁻⁴Vm 
=640 Vcm

Answer 2

The electric flux is given by the formula:

\[ \Phi_E = \vec{E} \cdot \vec{A} \]

where \( \vec{E} \) is the electric field and \( \vec{A} \) is the area vector. The area of one face of the cube is:

\[ A = (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]

Since the electric field is along the \( x \)-axis and the area vector is normal to the face of the cube, we have:

\[ \Phi_E = E \times A = 4000 \times (0.2)^2 = 4000 \times 0.04 = 640 \, \text{Vcm} \]

Thus, the electric flux is 640 Vcm.