Question

The electric flux from a cube of edge l is ϕ. If an edge of the cube is made \[2l\] and the charge enclosed is halved, its value will be :
A. \[{\mathbf{4}}\phi \]
B. \[{\mathbf{2}}\phi \]
C. \[\phi /{\mathbf{2}}\]
D. \[\phi \]

Answer 1

Hint:
- Apply Gauss law.
- You should know the definition of Electric flux.
- You should know the fundamental numerical values.

Step by step Solution:
Electric field is the region in which that force is felt. The electric field strength \[ = \] force per unit charge units \[ = \] newton’s per coulomb.it is also defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge. And Electric flux, property of an electric field that may be thought of as the number of electric lines of force \[\left( {or{\text{ }}electric{\text{ }}field{\text{ }}lines} \right)\] that intersect a given area.
Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.
In question edge of the cube is made \[2l\;\]and the charge enclosed is halved
\[l \to \dfrac{l}{2}\]
\[q \to \dfrac{q}{2}\]
By gauss’s law,
\[{\phi _1} = \dfrac{q}{{{\varepsilon _0}}}\]
\[{\phi _2} = \dfrac{q}{{2{\varepsilon _0}}}\] ,\[\phi \] is independent of dimensions
\[{\phi _2} = \dfrac{{{\phi _1}}}{2}\]
So, the Option (C) is correct.

Note:
- Try to apply laws in these kinds of problems.
- Concept should be thorough and clear.