Question

Compute the electric flux linked with a surface \( \vec{A} = 2 \hat{j} \, \mathrm{m}^2 \), placed in a uniform electric field \( \vec{E} = (4 \hat{i} + 3 \hat{j}) \, \mathrm{V/m} \).

Hint:

The electric flux is the dot product of the electric field vector and the area vector. If the vectors are perpendicular, the flux is zero; if parallel, the flux is the product of the magnitudes.

Answer 1

The electric flux \( \Phi_E \) linked with a surface is given by the dot product of the electric field \( \vec{E} \) and the area vector \( \vec{A} \):
\[ \Phi_E = \vec{E} . \vec{A} \] The area vector \( \vec{A} \) is given as \( \vec{A} = 2 \hat{j} \, \text{m}^2 \), and the electric field is \( \vec{E} = 4 \hat{i} + 3 \hat{j} \, \text{V/m} \). The dot product of the electric field and the area vector is:
\[ \Phi_E = (4 \hat{i} + 3 \hat{j}) . (2 \hat{j}) \] Since the dot product of two perpendicular vectors is zero and the dot product of two parallel vectors is the product of their magnitudes, we get:
\[ \Phi_E = (4 \times 0) + (3 \times 2) = 6 \, \text{V} . \text{m}^2 \] Thus, the electric flux linked with the surface is \( \Phi_E = 6 \, \text{V} . \text{m}^2 \).