The electric field in a region is given by $$\vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3$$ N/C. The flux of the field through a rectangular surface parallel to x-z plane is 6.0 Nm$$^2$$C$$^{-1}$$. The area of the surface is _____ cm$$^2$$.

Solution :

Given electric field :

$$\vec{E} = (2\hat{i}+4\hat{j}+6\hat{k}) \times 10^3\text{ N/C}$$

Surface is parallel to x-z plane.

Therefore, area vector is along y-axis.

Hence, only y-component of electric field contributes to flux.

So,

$$E_y = 4 \times 10^3\text{ N/C}$$

Electric flux is given by :

$$\phi = \vec{E}\cdot\vec{A}$$

$$= E_y A$$

Given,

$$\phi = 6.0\text{ Nm}^2\text{C}^{-1}$$

Therefore,

$$6 = (4 \times 10^3)A$$

$$A = \frac{6}{4 \times 10^3}$$

$$= 1.5 \times 10^{-3}\text{ m}^2$$

Converting into cm\(^2\) :

$$A = 1.5 \times 10^{-3} \times 10^4$$

= 15cm$$^2$$.

Final Answer :

$$15$$