Electrostatics

Step 1: Recall the formula for the electric potential of a point charge. The electric potential at a distance from a point charge is given by: where is Coulomb's constant.
Step 2: Calculate the distance from the origin to the given point. The charge is at the origin (0,0,0) and the point is at (3a,4a,0). The distance is the magnitude of the position vector, calculated using the distance formula in 3D:
Step 3: Substitute the distance into the potential formula.

Step 1: Apply Gauss's Law for a point outside the sphere. For a spherically symmetric charge distribution, the electric field outside the distribution (at ) is the same as if all the charge were concentrated at a point charge at the center.
Step 2: State the formula for the electric field of a point charge. The electric field at a distance from a point charge is given by Coulomb's Law: where .
Step 3: Conclude the electric field for the sphere. Since the point is outside the conducting sphere, we can treat the sphere's charge as a point charge at its center. Therefore, the electric field at a distance is:

Step 1: Apply Gauss's Law for a point inside the shell. Consider a spherical Gaussian surface of radius (where is less than the radius of the shell) concentric with the charged shell. Gauss's Law states that the net electric flux through this surface is equal to the enclosed charge divided by :
Step 2: Determine the enclosed charge. For a spherical shell, all the charge resides on the surface of the shell. The Gaussian surface at radius inside the shell encloses no charge.
Step 3: Solve for the electric field. Since , the net flux must be zero. Due to the spherical symmetry, if the electric field were non-zero, it would have a constant magnitude on the Gaussian surface. The integral would be . For this to be zero, the electric field must be zero. Thus, the electric field is zero everywhere inside a uniformly charged spherical shell.

Step 1: Relate electric potential and electric field. The electric field is related to the electric potential by . In one dimension, this is .
Step 2: Apply the given condition. We are given that the potential is constant inside the sphere. If is constant, its derivative with respect to position must be zero. So, the electric field inside the conductor is zero.
Step 3: Relate electric field to charge distribution using Gauss's Law. Gauss's Law in differential form (Poisson's equation) is , where is the volume charge density. Since everywhere inside the sphere, its divergence must also be zero. This means there is no net charge at any point inside the volume of the conductor. Any net charge on a conductor must reside on its surface.

Step 1: Recall the properties of a conductor in electrostatic equilibrium. One of the key properties is that the electric potential is constant throughout the entire conductor, including its surface. This means the surface of a conductor is an equipotential surface.
Step 2: Relate electric field lines to equipotential surfaces. Electric field lines are always perpendicular to equipotential surfaces. If there were a component of the electric field parallel (tangential) to the surface, it would exert a force on the charges on the surface and cause them to move.
Step 3: Apply the condition of electrostatic equilibrium. "Electrostatic equilibrium" means that the charges are no longer moving. For the charges to be stationary, there must be no net force on them parallel to the surface. This implies that the tangential component of the electric field must be zero.
Step 4: Conclude the relationship. If the tangential component of the electric field is zero, the electric field vector at the surface must be entirely perpendicular to the surface. Note that the field is not zero on the surface unless there is no charge on the conductor at all.

A fundamental property of a conductor in electrostatic equilibrium is that the net electric field inside the conductor must be zero. If there were an electric field inside, it would exert a force on the free charges (electrons) within the conductor, causing them to move. This movement of charge would mean the conductor is not in equilibrium. Therefore, the charges rearrange themselves on the surface of the conductor in such a way that they cancel out any external electric field within the bulk of the conductor. Regardless of the strength of the electric field just outside the conductor, the field just inside is always zero.

Step 1: Recall the formula for the capacitance of a parallel-plate capacitor. The capacitance of a parallel-plate capacitor with plate area and separation , filled with vacuum, is: where is the permittivity of free space.
Step 2: Recall the effect of inserting a dielectric material. When a dielectric material with dielectric constant is inserted to completely fill the space between the plates, the permittivity of the space becomes .
Step 3: Write the formula for the new capacitance. The new capacitance with the dielectric slab is:
Step 4: Compare the new capacitance with the original capacitance. Thus, the capacitance increases by a factor of K. Since the dielectric constant is always greater than 1 for materials, the capacitance always increases.