Electric Potential Energy: Derivations

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 q₁ from infinity to r₁.
Next, we consider the work done in bringing q₂ to r₂. In this step, work is done not only against the external field E but also against the field due to q₁.

• Work done on q₂ against the external field = q₂.V(r₂)
• Work done on q₂ against the field due to q₁
  where  r₁₂ is the distance between q₁ and q₂. 
• By the superposition principle for fields, we add up the work done on q₂ against the two fields (E and that due to q₁):
  Work done in bringing q₂ to r₂

_{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 q₁ = +q and q₂ = -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 θ₀ to angle θ₁ 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 θ₀ = 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, r₁ and r₂ denote the position vectors of +q and -q.
• Now, the potential difference between positions r₁ and r₂ equals the work done in bringing a unit positive charge against the field from r₂ to r₁.
• The displacement parallel to the force is 2a cosθ. Thus, [V(r₁)-V (r₂)] = -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 θ₀ = 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(r₁)  - V (r₂)] = 0.

Electric Potential Energy of Charged Particles

➤ Electric Potential Energy of a System of Two Charged Particles

• The figure shows two + ve charges q₁ and q₂ separated by a distance r. The electrostatic interaction energy of this system can be expressed as work done in bringing charge q₂ from infinity to the given separation from q₁.
• 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 Sphere

➤ 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 of Uniformly Charged Ring, Rod, and Disc

➤ 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: