An electric field is given by E = (6 i + 5 j + 3 k) N/C. The electric flux through a surface area 30 i m^2 lying in the YZ-plane (in SI units) is:

Given: E = (6 i + 5 j + 3 k) N/C and A = 30 i m^2. Find: Electric flux through the given surface. Electric flux is given by the dot product of the electric field vector and the area vector: = E A Since the surface lies in the YZ-plane, its area vector is along the i direction. Therefore, = (6 i + 5 j + 3 k) (30 i) Only the components along the same direction contribute in the dot product, so = 6 30 = 180 N m^2/ C Therefore, the electric flux through the surface is 180 N m^2/ C. The correct option is C.

An electric field is given by E = (6 i + 5 j + 3 k) N/C. The elec… - Solution | JEE Physics

MCQEasyJEE 2024Electric Field & Field Lines

An electric field is given by $\vec{E} = (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C}$ . The electric flux through a surface area $30\hat{i} \, \text{m}^2$ lying in the YZ-plane (in SI units) is:

A
$90$
B
$150$
C
$180$
D
$60$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $\vec{E} = (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C}$ and $\vec{A} = 30\hat{i} \, \text{m}^2$ .

Find: Electric flux through the given surface.

Electric flux is given by the dot product of the electric field vector and the area vector:

\Phi = \vec{E} \cdot \vec{A}

Since the surface lies in the YZ-plane, its area vector is along the $\hat{i}$ direction. Therefore,

\Phi = (6\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (30\hat{i})

Only the components along the same direction contribute in the dot product, so

\Phi = 6 \times 30

\Phi = 180 \, \text{N m}^2/\text{C}

Therefore, the electric flux through the surface is $180 \, \text{N m}^2/\text{C}$ . The correct option is C.

Common mistakes

Using the full magnitude of $\vec{E}$ instead of taking the dot product is incorrect because electric flux depends only on the component of the field along the area vector. Use $\Phi = \vec{E} \cdot \vec{A}$ , not $|\vec{E}|\,|\vec{A}|$ unless the vectors are parallel.
Taking the area vector in the wrong direction is incorrect because a surface in the YZ-plane has its normal along the $x$ -axis, that is, along $\hat{i}$ . Do not use $\hat{j}$ or $\hat{k}$ for the area vector here.
Adding all components $6 + 5 + 3$ is wrong because only the component parallel to $\hat{i}$ contributes to the flux through this surface. Use only the $6\hat{i}$ term while evaluating the dot product.

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