Question 62: Current Electricity

Question

The electric flux from cube of side $$1 \mathrm{~m}$$ is '$$\Phi$$' When the side of the cube is made $$3 \mathrm{~m}$$ and the charge enclosed by the cube is made one third of the original value, then the flux from the bigger cube will be :

Accepted Answer

The electric flux ($$\Phi$$) through a closed surface, such as the surface of a cube, is directly proportional to the charge enclosed ($$Q$$) by that surface. This relationship is dictated by Gauss's Law, which is represented by the equation:
$$\Phi = \frac{Q}{\varepsilon_0}$$
where:
$$\Phi$$ is the electric flux,
$$Q$$ is the charge enclosed by the surface,
$$\varepsilon_0$$ is the permittivity of free space (a constant).
In the given problem,
The initial flux from a cube of side $$1 \, 	ext{m}$$ is $$\Phi$$.
When the side of the cube is increased to $$3 \, 	ext{m}$$, and the charge enclosed is reduced to a third of its original value ($$Q' = \frac{Q}{3}$$), the relationship above can be used to find the new flux ($$\Phi'$$).
Using the relation for electric flux and substituting the new charge value, we get:
$$\Phi' = \frac{Q'}{\varepsilon_0} = \frac{\frac{Q}{3}}{\varepsilon_0} = \frac{1}{3} \cdot \frac{Q}{\varepsilon_0}$$
Since the original flux $$\Phi = \frac{Q}{\varepsilon_0}$$, substituting this back into the equation for $$\Phi'$$ yields:
$$\Phi' = \frac{1}{3} \cdot \Phi$$
Thus, the electric flux from the larger cube with side $$3 \, 	ext{m}$$ and one third the original enclosed charge is $$\frac{\Phi}{3}$$. Therefore, the correct answer is:
Option A: $$\frac{\Phi}{3}$$

Answer

$$\frac{\Phi}{3}$$

Answer

$$\Phi$$

Answer

$$3 \Phi$$

Answer

$$9 \Phi$$

Current Electricity

The electric flux from cube of side $1 \mathrm{~m} is '\Phi' When the side of the cube is made 3 \mathrm{~m}$ and the charge enclosed by the cube is made one third of the original value, then the flux from the bigger cube will be :

Choose the correct answer: