Electrostatics - Quiz

The dielectric constant tends to be high in insulating materials, such as ceramics and plastics. Insulators are materials that have very low electrical conductivity, which means they do not allow the flow of electric current easily. This characteristic makes them good dielectric materials with high dielectric constants. Insulators have a large number of bound electrons, and when an electric field is applied, these electrons get polarized, resulting in the accumulation of electric charge within the material. The ability of insulators to store a significant amount of electric charge leads to high dielectric constants.

The susceptibility (χe) can be calculated by dividing the bound charge density by the free charge density. In this case, the ratio is 12/6 = 2. Therefore, the susceptibility is 2. Susceptibility measures the extent to which a material can become polarized when exposed to an electric field.

The susceptibility of free space or air is zero (0). In free space, the relative permittivity (εr) is equal to 1, and the susceptibility (χe) is given by χe = εr - 1 = 1 - 1 = 0. This means that free space does not exhibit any polarization or ability to become polarized under the influence of an electric field.

The electric potential at a point due to a point charge q is given by V = q/r, where r is the distance between the charge and the point. Therefore, the electric potential due to a point charge q at a distance r in the air is qr.

The electric field intensity E at a distance r from a point charge q is given by E = k*q/r^2, where k is the electrostatic constant. The electric potential V at that distance is given by V = k*q/r. Given that E = 32 N/C and V = 16 J/C, we can equate the two expressions and solve for r. Therefore, 32 N/C = (16 J/C) / r. Rearranging the equation, we have r = (16 J/C) / 32 N/C = 0.5 m.

The potential at the midpoint between the two equal and opposite charges is 0 V. Since the charges have the same magnitude but opposite signs, their individual potentials cancel each other out, resulting in a net potential of 0 V.

Electric field lines generated radially from a positive point charge, and equipotential surfaces are always perpendicular to electric field lines. In order for the equipotential surfaces to be perpendicular to the electric field lines and encompass all directions around the charge, they must be spherical in shape.

In a charged conductor, the charges distribute themselves uniformly on its surface. This distribution of charges creates an electric field inside the conductor. However, within a conductor, the electric field is zero, and therefore, the potential is constant throughout its surface, making it an equipotential surface. The electric field lines are perpendicular to the surface of the conductor.

A positive charge will move from a region of higher potential to a region of lower potential. Since the potential decreases as we move away from the positive point charge, the charge placed between A and B will experience a force that pushes it towards B, causing it to move from A to B.

In order for charge to flow from the smaller sphere to the larger sphere, the spheres should be placed concentrically, with the smaller sphere positioned inside the larger sphere. When connected with a wire, the charge will redistribute itself on the outer surfaces of both spheres, with some charge flowing from the smaller sphere to the larger sphere.