JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 1)
A cube is placed inside an electric field, $$\overrightarrow E = 150{y^2}\widehat j$$. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :
_1st_September_Evening_Shift_en_1_1.png)
3.8 $$\times$$ 10$$-$$11 C
8.3 $$\times$$ 10$$-$$11 C
3.8 $$\times$$ 10$$-$$12 C
8.3 $$\times$$ 10$$-$$12 C
Explanation
Given, the side of the cube, s = 0.5 m
Electric field, E = 150 y2$$\widehat j$$
The direction of electric field is as shown in the below figure,
_1st_September_Evening_Shift_en_1_2.png)
At bottom surface, y = 0
As we know that, the expression of electric flux,
$$\phi$$ = E . A cos$$\theta$$
Here, E is the electric field passing through the cube and A is the surface area of the cube.
Substituting the values in the above equations, we get
$$\phi$$ = 150y2 . (0.5 $$\times$$ 0.5) $$\times$$ cos180$$^\circ$$
= 150(0)2 . (0.25) $$\times$$ ($$-$$1) = 0
Hence, the electric flux is zero at the bottom surface.
At the top surface, y = 0.5 m
Electric field, E = 150 y2$$\widehat j$$ = 150(0.5)2 = 37.5 N/C
Electric flux at the top surface,
$$\phi$$ = E . A cos$$\theta$$
= (37.5) . (0.5 $$\times$$ 0.5) cos0$$^\circ$$
= 9.375 N / C - m2
By using the Gauss's law, $$\phi = {{{Q_{in}}} \over {{\varepsilon _0}}}$$
Here, Qin = net charge enclosed in the cube and $${{\varepsilon _0}}$$ = permittivity of the free space.
Substituting the values in the above equation, we get
$$9.375 = {{{Q_{in}}} \over {8.85 \times {{10}^{ - 12}}}}$$
Qin = 8.3 $$\times$$ 10-11 C
The charge inside the cube is 8.3 $$\times$$ 10-11 C.
Electric field, E = 150 y2$$\widehat j$$
The direction of electric field is as shown in the below figure,
At bottom surface, y = 0
As we know that, the expression of electric flux,
$$\phi$$ = E . A cos$$\theta$$
Here, E is the electric field passing through the cube and A is the surface area of the cube.
Substituting the values in the above equations, we get
$$\phi$$ = 150y2 . (0.5 $$\times$$ 0.5) $$\times$$ cos180$$^\circ$$
= 150(0)2 . (0.25) $$\times$$ ($$-$$1) = 0
Hence, the electric flux is zero at the bottom surface.
At the top surface, y = 0.5 m
Electric field, E = 150 y2$$\widehat j$$ = 150(0.5)2 = 37.5 N/C
Electric flux at the top surface,
$$\phi$$ = E . A cos$$\theta$$
= (37.5) . (0.5 $$\times$$ 0.5) cos0$$^\circ$$
= 9.375 N / C - m2
By using the Gauss's law, $$\phi = {{{Q_{in}}} \over {{\varepsilon _0}}}$$
Here, Qin = net charge enclosed in the cube and $${{\varepsilon _0}}$$ = permittivity of the free space.
Substituting the values in the above equation, we get
$$9.375 = {{{Q_{in}}} \over {8.85 \times {{10}^{ - 12}}}}$$
Qin = 8.3 $$\times$$ 10-11 C
The charge inside the cube is 8.3 $$\times$$ 10-11 C.
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