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 Charged Particles➤ 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 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: