Question

A surface element ds=5i is placed in an electric field e=4i+4j+4k what is electric flux

Answer 1

Answer

Here, ds→=5iˆ,E→=4iˆ+4jˆ+4kˆ

Electric flux, ϕ=E→.ds→=(4iˆ+4jˆ+4kˆ)⋅5iˆ

ϕ=20units

Answer 2

Given:

• Surface element ds=5i

• Field intensity E = 4i + 4j + 4k.

To find:

• Electric flux ?

Calculation:

The electric flux can be calculated by a scalar product (also known as DOT PRODUCT) between the field intensity vector and the surface area vector.

• Let flux be $\phi$:

• Also, we should know that dot product between perpendicular vectors is zero.

$\phi = \vec{E} \: . \: \vec{ds}$

$\implies \: \phi = (4 \hat{i} + 4 \hat{j} + 4 \hat{k}).(5 \hat{i})$

$\implies \: \phi = (4 \times 5) + (0) + (0)$

$\implies \: \phi = (4 \times 5)$

$\implies \: \phi = 20 \: N {m}^{2} /C$

So, the electric flux is 20 Nm²/C.