Question 481: Electrostats & Capacitance

Question

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 :

Accepted Answer

Given, the side of the cube, s = 0.5 m
Electric field, E = 150 y
2
$$\widehat j$$
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$$	heta$$
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
$$\phi$$ = 150y
2
. (0.5 $$	imes$$ 0.5) $$	imes$$ cos180$$^\circ$$
= 150(0)
2
. (0.25) $$	imes$$ ($$-$$1) = 0
Hence, the electric flux is zero at the bottom surface.
At the top surface, y = 0.5 m
Electric field, E = 150 y
2
$$\widehat j$$ = 150(0.5)
2
= 37.5 N/C
Electric flux at the top surface,
$$\phi$$ = E . A cos$$	heta$$
= (37.5) . (0.5 $$	imes$$ 0.5) cos0$$^\circ$$
= 9.375 N / C - m
2
By using the Gauss's law, $$\phi  = {{{Q_{in}}} \over {{\varepsilon _0}}}$$
Here, Q
in
= net charge enclosed in the cube and $${{\varepsilon _0}}$$ = permittivity of the free space.
Substituting the values in the above equation, we get
$$9.375 = {{{Q_{in}}} \over {8.85 	imes {{10}^{ - 12}}}}$$
Q
in
= 8.3 $$	imes$$ 10
-11
C
The charge inside the cube is 8.3 $$	imes$$ 10
-11
C.

Answer

3.8 $$	imes$$ 10$$-$$11 C

Answer

8.3 $$	imes$$ 10$$-$$11 C

Answer

3.8 $$	imes$$ 10$$-$$12 C

Answer

8.3 $$	imes$$ 10$$-$$12 C

Electrostats & Capacitance

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 :

Choose the correct answer: