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**Electric Potential Energy in An External Field**

➤ **Potential Energy of a Single Charge**

Potential energy of q at r in an external field:

U = q.V(r),

where V(r) is the external potential at the point r.

➤ **Potential Energy of a System of two Charges in an External Field**

First, we calculate the work done in bringing the charge q_{1} from infinity to r_{1}.

Next, we consider the work done in bringing q_{2} to r_{2}. In this step, work is done not only against the external field E but also against the field due to q_{1}.

- Work done on q
_{2}against the external field = q_{2}.V(r_{2}) - Work done on q
_{2}against the field due to q_{1}

where r_{12}is the distance between q_{1}and q_{2}. - By the superposition principle for fields, we add up the work done on q
_{2}against the two fields (E and that due to q_{1}):

Work done in bringing q_{2}to r_{2}_{}

_{Thus, }_{Potential energy of the system }_{= Total work done in assembling the configuration}_{⇒ Potential Energy of the System}

➤ **Electric ****Potential Energy of a Dipole in an External Field**

- Consider a dipole with charges q
_{1}= +q and q_{2}= –q placed in a uniform electric field E, as shown in Figure below. - In a uniform electric field, the dipole experiences no net force; but experiences a torque τ given by τ = p × E, which will tend to rotate it (unless p is parallel or anti-parallel to E).
- Suppose an external torque τ
_{ext}is applied in such a manner that it just neutralizes this torque and rotates it in the plane of paper from angle θ_{0}to angle θ_{1}at an infinitesimal angular speed and without angular acceleration. __The amount of work done by the external torque will be given by:__- This work is stored as the potential energy of the system. We can then associate potential energy U (θ) with an inclination θ of the dipole.
- Similar to other potential energies, there is freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take θ
_{0}= π / 2. (Αn explanation for it is provided towards the end of discussion) - We can then write,
- On comparing with the case of System of two Charges in an External Field.

The potential energy expression is:

Here, r_{1}and r_{2}denote the position vectors of +q and –q. - Now, the potential difference between positions r
_{1}and r_{2}equals the work done in bringing a unit positive charge against the field from r_{2}to r_{1}. - The displacement parallel to the force is 2a cosθ. Thus, [V(r
_{1})–V (r_{2})] = –E × 2a cosθ.

__We thus obtain:__

- We note that U′ (θ) differs from U (θ) by a quantity which is just a constant for a given dipole. We can now understand why we took θ
_{0}= π/2. In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V(r_{1}) – V (r_{2})] = 0.

**➤ Electric Potential Energy of a System of Two Charged Particles**

- The figure shows two + ve charges q
_{1}and q_{2}separated by a distance r. The electrostatic interaction energy of this system can be expressed as work done in bringing charge q_{2}from infinity to the given separation from q_{1}. __It can be calculated as:__

, where – ve sign shows that x is decreasing.__If the two charges are of__**opposite signs**, then potential energy will be negative as:

**➤ Electric ****Potential Energy for a System of Multiple Charged Particles**

- When more than two charged particles are there in a system, the interaction energy can be given as the
**sum of interaction energies of all the different possible pairs of particles**. **Example:**If a system of three particles having charges q1, q2 and q3 is given as shown in figure.__The total interaction energy of this system can be given as:__

**➤ Electric Potential Due to a Conducting Sphere or a Shell**

__Outside the Sphere__: According to definition, Electric potential at the point P__On the Sphere:____Inside the Sphere:__

∵ Inside the surface E = 0 , dV/dr = 0

➤ **Electric Potential Due to Solid Non-Conducting Sphere**

__Outside the sphere__: Same as conducting sphere.__On the Surface__: Same as conducting sphere.__Inside the sphere__:

➤ **Electric Potential Due to a Uniformly Charged Rod**

Electric Potential due to a uniformly charged rod of length L and linear charge density lambda, at a point P on its axial line which is d units away from it.Charge per unit length, λ = Q/L

Charge of slice, dq = λ.dx

__Thus, Electric potential due to uniformly charged rod:__

➤ **Electric Potential Due to a Charged Ring**

Let electric potential at point P due to small element of length dℓ be dV.

(distance between small element and point P is equal to r)__Thus, Electric potential due to whole ring:__

➤ **Potential Due to a Uniformly Charged Disc**

Find the electric potential at the axis of a uniformly charged disc and use potential to find the electric field at same point.__Thus, Electric Potential due to ring element of radius r and thickness dr is dv:__

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