Electrostatics

Step 1: The capacitance of a capacitor with a dielectric is given by , where is the dielectric constant.
Step 2: With the dielectric, the capacitance is , where is the original capacitance without the dielectric.
Step 3: Removing the dielectric, the capacitance returns to . Thus, .
Step 4: Therefore, the new capacitance is . \bigskip

The electrostatic force between two charges is governed by Coulomb's law: where and are the magnitudes of the charges and is the distance between them. When a charge is placed on a spherical conductor, the electrostatic force is maximum when the charge is placed at the surface because the charge distributes uniformly over the surface of the sphere. This results in the maximum electrostatic force at the surface. Thus, the electrostatic force is maximum when the charge is placed at the surface of the sphere.

According to Gauss's law, the total charge enclosed by a surface is proportional to the net electric flux through the surface. Mathematically, Gauss's law is expressed as: Where: - is the total electric flux, - is the charge enclosed, - is the permittivity of free space. The total flux is the sum of the inward and outward flux: Thus, the total charge enclosed is:

Step 1: Understanding the problem.}
We are given three charges: at position 0, at position , and at position . The net force on charge is zero, so we need to find the value of that satisfies this condition.
Step 2: Forces on charge .}
The force on charge due to charge (placed at the origin) and charge (placed at ) must balance each other. The force due to is repulsive, and the force due to is attractive if is negative.
Step 3: Apply Coulomb's Law.}
The force between and is: The force between and is: For the forces to balance, we set these equal: Simplifying: Thus, the value of is .

Step 1: Review of the Problem Setup.}
The problem involves the electric field inside a uniformly charged spherical shell. We know that the electric field produced by a charge distribution is determined by Gauss’s law. The spherical symmetry of the shell implies that the electric field at any point inside the shell should only depend on the geometry of the shell, not the distribution of charge itself.
Step 2: Applying Gauss’s Law.}
Gauss’s law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is written as: where is the charge enclosed by the surface, and is the permittivity of free space.
Step 3: Choosing a Gaussian Surface Inside the Shell.}
Now, consider a spherical Gaussian surface inside the shell. By symmetry, the electric field should point radially outward and be uniform over the spherical surface. Since the shell is uniformly charged, the charge enclosed by the Gaussian surface is zero. The reason is that there is no charge enclosed within the spherical shell (the charge resides only on the shell, not inside it).
Step 4: Consequence of Gauss’s Law.}
Since , from Gauss’s law, the total electric flux through the Gaussian surface must also be zero. The electric flux is the integral of the electric field over the surface area, and since the flux is zero, the electric field at any point inside the shell must also be zero.
Step 5: Final Conclusion.}
Therefore, the electric field inside a uniformly charged spherical shell is zero. This result holds for any point inside the shell, no matter how close or far it is to the center.

Step 1: Concept of electric field due to point charges.}
The electric field at any point due to a point charge is given by: where is Coulomb's constant, is the charge, and is the distance from the charge to the point where the field is being calculated.
Step 2: Condition for zero electric field.}
For the electric field to be zero at some point between two charges, the electric fields due to both charges must be equal in magnitude and opposite in direction. Let the distance from charge to the point where the electric field is zero be . The distance from charge to the same point is cm, since the total distance between the charges is 15 cm.
Step 3: Setting up the equation.}
The magnitudes of the electric fields due to the charges must satisfy: Substitute the values and : Cross-multiply to solve for : Simplifying: Expanding both sides: Divide the equation by 3: Solve this quadratic equation using the quadratic formula: Thus, or . Since distance cannot be negative, we have cm.
Step 4: Final answer.}
Therefore, the distance from to the point where the electric field is zero is 5 cm from .