Electrostatics

Given: Two concentric conducting spheres with radii and . The inner sphere is given a charge , and the outer sphere is grounded (i.e., its potential is zero). We are to find the potential at a point where , which lies between the two spheres.

Concept: When the outer sphere is grounded, it induces a charge on its inner surface (to neutralize the electric field outside), and a charge appears on its outer surface to maintain neutrality. The potential at any point between the spheres (i.e., for ) is due to both the inner sphere and the induced charge on the inner surface of the outer sphere.

Step-by-step calculation: Potential at a point due to a spherical shell of charge is: Let us consider: - Inner sphere at radius has charge - Induced charge appears on the inner surface of the outer sphere at radius So at , total potential is the sum of potentials due to: 1. Charge at center (inner sphere): 2. Charge at radius : Since the field inside a shell is same as if all the charge were at center, potential due to shell at point inside is:

Total Potential:

Final Answer:

Step 1: Understand Electric Flux

Electric flux, , through a surface is defined as: For a uniform electric field and a flat surface, this simplifies to: where is the area of the surface and is the angle between the electric field and the normal vector to the surface.

Step 2: Flux Through the Circular Surface

The circular surface is perpendicular to the electric field. Therefore, the angle between the electric field and the normal to the surface is 180 degrees (or 0 degrees, depending on which direction you define the normal to point). The area of the circular surface is . The flux through the circular surface is: The negative sign indicates that the flux is entering the hemisphere.

Step 3: Flux Through the Curved Surface

Since there is no charge enclosed within the hemisphere, according to Gauss's Law, the total flux through the closed surface (hemisphere) must be zero. Therefore, the flux through the curved surface must be equal in magnitude but opposite in sign to the flux through the circular surface: This is the outgoing flux.

Step 4: Total Flux Enclosed

As mentioned in Step 3, Gauss's Law states that the total electric flux through a closed surface is proportional to the enclosed charge. Since there is no charge enclosed within the hemisphere, the total flux enclosed is zero:

Step 5: Work Done in Moving a Charge

The work done in moving a point charge from point to point in an electric field is given by: where is the potential difference between points and . In a uniform electric field, the potential difference only depends on the displacement along the direction of the electric field. Since points A and B are at the same height, the electric potential difference is independent of R. Therefore, the work done depends on the potential difference between points and , and it is independent of the path taken (since the electric field is conservative) and also independent of .

Step 6: Determine the Correct Statements

1. Flux through the curved surface of hemisphere is πR²E: This statement is TRUE.
2. Flux through the circular surface of hemisphere is πR²E: This statement is FALSE; the flux is -πR²E.
3. Total flux enclosed is zero: This statement is TRUE.
4. Work done in moving a point charge q from A to B via the path ACB depends upon R: This statement is FALSE.

Conclusion

The correct statements are:

• Flux through the curved surface of the hemisphere is πR²E.
• Total flux enclosed is zero.

Given: A neutral conducting solid sphere with two spherical cavities. The radii of the cavities are and , and the center-to-center distance between the two cavities is . Charges and are placed at the centers of the cavities respectively.

To find: The force between the charges and .

Key Concept: In a conductor, charges redistribute themselves on the outer surface in such a way that the electric field inside the conductor is zero. Furthermore, the presence of cavities in the conductor does not affect the electric field within the conducting material itself. The force between the charges in the cavities is influenced by the conducting nature of the sphere and its symmetry.

Solution: Since the conductor is neutral, the electric field inside the conductor is zero, and the charges on the cavities do not directly interact with each other in terms of electrostatic force. The conducting sphere ensures that the electric fields created by and do not interact in the usual way. Thus, the force between the charges and is:

Final Answer: The force between the charges and is zero.

Given: Two charges, each equal to , are placed at and . A charge is placed at the origin. We are asked to find the force acting on the charge when it undergoes a small displacement along the -direction.

Approach: The force on a charge in the presence of other charges is given by Coulomb's Law: where: - is Coulomb's constant, - and are the magnitudes of the charges, - is the distance between the charges. In this case, the two charges at and exert forces on the charge placed at the origin. When the charge is displaced slightly along the -direction, the distance between and the charges at and changes. The displacement introduces small asymmetry in the forces from the two charges, which can be calculated using Coulomb's Law and considering the vector nature of the force.

Resulting Force: After analyzing the force components, we find that the force on due to a small displacement along the -direction is proportional to the displacement , with a negative sign indicating that the force is directed toward the origin (restoring force).

Conclusion: The force acting on when it is displaced along the -direction is proportional to .

Final Answer: The force is proportional to .