Electrostatic Potential And Capacitance

An electrical capacitor is a device that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material (called the dielectric). When a potential difference is applied across the plates, a charge is stored on each plate.
The capacitance of a parallel plate capacitor is given by the expression: Where:
- is the capacitance,
- is the permittivity of free space ( ),
- is the area of one of the plates,
- is the separation between the plates.
The capacitance depends on the following factors:
1. The area of the plates: A larger area increases the capacitance.
2. The separation between the plates: A smaller distance between the plates increases the capacitance.
3. The dielectric material between the plates: The dielectric constant ( ) of the material increases the capacitance. The capacitance with a dielectric is given by: Where is the dielectric constant of the material.

Let the capacitance of each capacitor be and the total voltage of the battery is . Since the capacitors are connected in series, the equivalent capacitance of the series combination is given by: Thus, the equivalent capacitance is: Now, using the relation for voltage and charge in a capacitor, the potential difference across each capacitor is: Therefore, the potential at point is 20 V (since it is connected directly to the battery) and the potential at point is 0 V (since it is earthed). Thus, the potential at is 20 V and at is 0 V.

The electric potential energy of a system of point charges is the work done to assemble the system of charges from infinity. It is given by the formula:
where:
- is the total electric potential energy of the system,
- is Coulomb's constant,
- and are the charges,
- is the distance between charges and .
In the given figure, we have a system of four point charges arranged at the corners of a square. The charges are:
- at ,
- at ,
- at ,
- at .
The distances between each pair of charges are either (side of the square) or (diagonal of the square).
Step 1: Calculate the potential energy due to the interactions between charges on adjacent sides (distance ):
- Interaction between and at and : - Interaction between and at and : - Interaction between and at and : - Interaction between and at and : Thus, the total potential energy due to adjacent charges is: Step 2: Calculate the potential energy due to the interactions between charges on the diagonals (distance ):
- Interaction between at and at : - Interaction between at and at : Thus, the total potential energy due to diagonal charges is: Step 3: Total electric potential energy of the system: Now, the total electric potential energy of the system is the sum of the contributions from adjacent and diagonal interactions: Thus, the total electric potential energy of the system is: This is the total electric potential energy of the system of point charges arranged in a square.

The surface charge density is the charge per unit area on the surface of a conductor. It is given by the formula:
Where:
- is the charge distributed over the surface,
- is the surface area of the sphere.
For a sphere, the surface area is given by:
Thus, the surface charge density for a sphere of radius is:
The unit of surface charge density is:
Now, we are given two concentric hollow spheres with radii and , and their surface charge densities are equal. This implies:
Thus, we have the following equations for the surface charge densities of both spheres:
Since , we can equate these expressions:
This simplifies to:
However, this is not possible if , implying that the surface charge densities cannot be the same unless . Electric Potential at the Common Centre Now, let's calculate the electric potential at the common centre (which is the centre of both spheres). The electric potential at the centre of a spherical shell due to a charge distributed uniformly on its surface is given by:
For a concentric configuration of two spheres, the potential at the common centre will be the sum of the potentials due to both spheres. Therefore, the total potential at the common centre is:
Thus, the electric potential at the common centre of the spheres is:
Where: - is the charge distributed on the spheres,
- and are the radii of the spheres,
- is the permittivity of free space.

Capacitor Combination: The given combination of capacitors is shown below:
1. Step 1: Calculate the Equivalent Capacitance of the 4 μF and 5 μF Capacitors in Series:}
For capacitors in series, the equivalent capacitance is given by: Substituting the values: Thus: 2. Step 2: Find the Equivalent Capacitance of the Entire Combination:}
Now, the equivalent capacitance of the 3 μF capacitor in parallel with the series combination (2.22 μF) is: So, the total equivalent capacitance of the combination is 5.22 μF.
3. Step 3: Calculate the Total Energy Stored:}
The energy stored in a capacitor is given by the formula: Substituting the values ( and ): Thus, the total energy stored in the combination is 10.44 μJ.

In the given circuit, we have four capacitors , , , and connected in a combination of series and parallel. The total voltage across the combination is given as . Let's break down the solution into steps:
Step 1: Simplifying the Combination of Capacitors
First, we need to combine capacitors , , and in parallel. The total capacitance for parallel combination is given by: Substituting values: Step 2: Combining with in Series
Now, we combine the equivalent capacitance with in series. The equivalent capacitance for capacitors in series is given by: Substitute the values: Thus, the equivalent capacitance of the combination is approximately .
Step 3: Charge on Capacitor
The charge on the capacitor is given by: Substitute the values: So, the charge on capacitor is .
Step 4: Total Stored Energy
The total stored energy in the combination of capacitors is given by: Substitute the values: Thus, the total stored energy is .

The electrical capacity, or capacitance, of a conductor refers to its ability to store electric charge when a potential difference is applied across it. The capacitance of a conductor is defined as the amount of charge it can store per unit potential difference . Mathematically, it is given by the equation:
Where:
- is the capacitance in farads (F),
- is the charge stored on the conductor in coulombs (C),
- is the potential difference across the conductor in volts (V).
The unit of capacitance is the farad (F), which is defined as the amount of charge that can be stored per unit potential difference. In terms of SI units, we can express the farad as:
This means that if a conductor has a capacitance of 1 farad, it will store 1 coulomb of charge when a potential difference of 1 volt is applied across it.
The electrical capacity (or capacitance) depends on the physical characteristics of the conductor, such as its surface area, shape, and the dielectric properties of the material surrounding it. For a parallel plate capacitor, for example, the capacitance is given by:
Where:
- is the capacitance,
- is the permittivity of free space ( ),
- is the area of the plates,
- is the distance between the plates.
Thus, the electrical capacity of a conductor is the ratio of the charge stored to the potential difference, and its unit is the farad (F).

An equipotential surface is a surface on which the electric potential is constant at every point. It is defined such that no work is done in moving a charge along the surface because the electric potential difference between any two points on the surface is zero. The electric field is always perpendicular to the equipotential surface.
To answer the question, we need to consider the implications if two equipotential surfaces were to intersect.
1. Electric Potential at a Point:
- The electric potential at a point in space is a scalar quantity, meaning that at any given point, the potential has a single value.
- If two equipotential surfaces were to intersect, it would imply that a point would have two different potential values (since each surface corresponds to a different potential). This is a contradiction because a point cannot have two distinct potential values at the same time.
2. Electric Field Direction:
- The electric field is always perpendicular to the equipotential surface. If two equipotential surfaces were to intersect, the direction of the electric field would have to be ambiguous at the point of intersection because there would be two perpendicular directions corresponding to the two surfaces. This would violate the definition of the electric field.
3. Conclusion:
- Therefore, two equipotential surfaces can never intersect. If they did, it would imply a violation of the uniqueness of electric potential at any given point in space, and it would lead to an undefined situation for the electric field.
Thus, two equipotential surfaces cannot intersect.

Part 1: Deduction of Potential Energy of a Capacitor

Step 1: Understanding the Concept:
The potential energy stored in a capacitor is equal to the total work done in charging the capacitor. This work is done by an external agent (like a battery) to transfer charge from one plate to another against the electrostatic forces.

Step 2: Derivation:
Let's consider a capacitor of capacitance . At an intermediate stage during charging, let be the charge on the plates and be the potential difference between them. By definition, .
The work done ( ) to transfer an additional infinitesimal charge from the negative plate to the positive plate is given by: Substituting : To find the total work done ( ) in charging the capacitor from an initial charge of 0 to a final charge of , we integrate : Since is a constant: This work done is stored as electrostatic potential energy ( ) in the capacitor. Using the relation , we can express the energy in other forms: \begin{itemize} \item \item \end{itemize} Thus, the potential energy is .
Part 2: Derivation of Energy Density

Step 1: Understanding the Concept:
Energy density ( ) is the electrostatic potential energy stored per unit volume in the electric field.

Step 2: Derivation:
Consider a parallel plate capacitor with plate area and separation distance . The volume of the space between the plates is .
The capacitance of this capacitor is given by: The electric field between the plates is uniform and is related to the potential difference by: Now, let's use the potential energy formula and substitute the expressions for and : Energy density ( ) is the energy per unit volume: This shows that the energy density in the electric field between the plates of a charged capacitor is .

Step 1: Understanding the Concept:
A parallel plate capacitor consists of two large, flat, parallel conducting plates, each with an area A, separated by a small distance d. Capacitance (C) is a measure of its ability to store electric charge, defined as the ratio of the charge (Q) on one plate to the potential difference (V) between the plates.

Step 2: Derivation of the Expression:
Let the two plates of the capacitor have a surface charge density of and . The charge on each plate is .
Assuming the distance d is small compared to the dimensions of the plates, the electric field ( ) between them is uniform and is given by: where is the permittivity of free space.
The potential difference ( ) between the plates is the work done in moving a unit charge from one plate to another, which is . The capacitance C is defined as . Substituting the expression for V: This is the expression for the capacitance of a parallel plate capacitor with a vacuum (or air) between the plates.
If a dielectric material with a dielectric constant K is inserted between the plates, the capacitance becomes:

Step 3: How to Increase the Capacity:
Based on the formula , the capacitance can be increased in the following ways: \begin{enumerate} \item By increasing the area of the plates (A): Capacitance is directly proportional to the area of the plates. \item By decreasing the distance between the plates (d): Capacitance is inversely proportional to the separation between the plates. \item By introducing a dielectric medium: Inserting a dielectric material with a high dielectric constant (K > 1) between the plates increases the capacitance by a factor of K. \end{enumerate}