Electric Charges And Fields

Step 1: The electric field is related to the potential by the relation: Given , the electric field is: Step 2: The charge enclosed by a surface can be calculated using Gauss's Law: Since the electric field varies with , we integrate over the volume of the cube.
Step 3: After performing the integration, the charge enclosed in the cube is:

Step 1: If the electric and magnetic fields are parallel, the force on the charged particle due to the electric field will be in the same direction as the magnetic force.
Step 2: Since both forces act in the same direction, the charged particle will experience a uniform force, causing it to accelerate in a straight line.

Step 1: According to Gauss's Law, the total electric flux through a closed surface is given by: where is the enclosed charge and is the permittivity of free space.
Step 2: For a hemispherical surface, if the charge is placed at the center, it will be symmetrically distributed. The flux passing through the entire hemispherical surface will be .
Step 3: Since the flux is symmetrically distributed over the surface, half of the flux will pass through the curved surface, and the other half will pass through the flat surface.
Step 4: The flux passing through the flat surface is zero because the flat surface is in the opposite direction to the field lines, and the flux passing through the flat surface cancels out.

The electric flux through a surface is given by: where: - is the electric field,
- is the infinitesimal area vector normal to the surface.
Step 1: Given Data
- The electric field is given as:
- The shaded area in the figure is a square of side , but it is inclined at to the x-axis.
Step 2: Area Vector and Flux Calculation: The area vector is normal to the surface. Since the plane is inclined at an angle of to the x-axis, the normal to the plane makes an angle of with the x-axis. Thus, the component of the area vector along the x-axis is: Since the area of the square is: we get: The flux is then calculated as:
Step 3: Correct Answer
Since the closest correct answer in the given options is , we must verify our initial interpretation of the diagram. If the normal to the surface is actually along the x-axis, then the full area contributes to the flux, giving: Thus, the correct answer is: