A cube is placed inside an electric field, $$\vec{E} = 150y^2\hat{j}$$. 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:

We need to determine the total electric charge enclosed inside a cube of side length $$a = 0.5\text{ m}$$ placed in a position-dependent electric field.

1. Analyze the Electric Field and Cube Orientation

From the diagram , the electric field vector is given as:

$$\vec{E} = 150 y^2 \hat{j}$$

This tells us two very important things:

• The electric field points exclusively along the $$y$$-axis ($$\hat{j}$$ component). Because there are no components along the $$\hat{i}$$ or $$\hat{k}$$ directions, the net electric flux through the four side faces parallel to the $$y$$-axis (left, right, front, and back faces) is exactly zero.
• The strength of the electric field depends entirely on the height coordinate $$y$$. Therefore, flux will only pass through the bottom face and the top face of the cube.

2. Calculate the Net Electric Flux ($$\Phi_{\text{net}}$$)

Let's evaluate the electric flux through the two active faces:

1. Bottom Face (located at $$y = 0$$):
   Substitute $$y = 0$$ into the electric field equation:

   $$\vec{E}_{\text{bottom}} = 150(0)^2 \hat{j} = 0$$

   Since the electric field strength is zero at the base plane, the flux passing through the bottom face is:

   $$\Phi_{\text{bottom}} = 0$$

2. Top Face (located at $$y = a = 0.5\text{ m}$$):
   The area of this square face is $$A = a^2 = (0.5)^2 = 0.25\text{ m}^2$$. The outward normal vector for the top surface points upwards ($$+\hat{j}$$).
   Substitute $$y = 0.5\text{ m}$$ into the electric field equation:

   $$\vec{E}_{\text{top}} = 150(0.5)^2 \hat{j} = 150 \times 0.25 \hat{j} = 37.5 \hat{j}\text{ N C}^{-1}$$

   The flux passing out of the top face is:

   $$\Phi_{\text{top}} = \vec{E}_{\text{top}} \cdot \vec{A} = 37.5 \times 0.25 = 9.375\text{ N m}^2\text{ C}^{-1}$$

The total net electric flux ($$\Phi_{\text{net}}$$) emerging from the entire cube is:

$$\Phi_{\text{net}} = \Phi_{\text{top}} + \Phi_{\text{bottom}} = 9.375 + 0 = 9.375\text{ N m}^2\text{ C}^{-1}$$

3. Find the Enclosed Charge ($q_{\text{in}}$) Using Gauss's Law

According to Gauss's Law, the net electric flux through any closed surface is proportional to the net charge enclosed inside it:

$$\Phi_{\text{net}} = \frac{q_{\text{in}}}{\varepsilon_0} \implies q_{\text{in}} = \Phi_{\text{net}} \cdot \varepsilon_0$$

Using the standard permittivity constant value ($$\varepsilon_0 = 8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$$):

$$q_{\text{in}} = 9.375 \times 8.854 \times 10^{-12}$$

$$q_{\text{in}} \approx 83.006 \times 10^{-12}\text{ C} = 8.3 \times 10^{-11}\text{ C}$$

Conclusion

The total net charge inside the cube is $$8.3 \times 10^{-11}\text{ C}$$, which corresponds exactly to Option A.