Important Derivations: Electric Charges and Fields | Physics Class 12 - NEET PDF Download

Important Derivations

Q1: Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
Sol:

Electric field intensity due to a thin infinite sheet of charge: Let σ be the surface density of charge and P be a point at a distance r from the sheet where  has to be calculated. on either side is perpendicular to the sheet.
Imagine a cylinder of cross-sectional area ds around P and length 2r, piercing through the sheet. At the two edges,
At the curved surfaces
So, there is no contribution to electric flux from the curved surfaces of the cylinder.

Electric flux over the edges

Total charge enclosed by the cylinder
By Gauss’s theorem,
From the expression, it is clear that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

Q2: Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field.
Sol:

Let the dipole moment  makes an angle θ with the direction of

Force on the charge -q = -qE
Force on the charge q = +qE
These forces are equal in magnitude but act along parallel lines at different points hence form a couple which tends to rotate the dipole in the anticlockwise direction.
As, torque = moment of force.

Q3: Derive an expression for the electric field due to an infinitely long straight wire of linear charge density λ C/m.
Sol:

Electric field due to an infinitely long straight wire:
Consider an infinitely long line charge having linear charge density λ. To determine its electric field at distance $r$, consider a cylindrical Gaussian surface of radius r and length ℓ coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S1 and is directed radially outward.

Total flux through the cylindrical surface,
As λ is the charge per unit length and ℓ is the length of the wire,
so charge enclosed,
By Gauss’s theorem,
Q4: Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
Electric field intensity at point P due to +q (at B)

Electric field intensity at P due to q charge  (at A )

Clearly,  in magnitude.
 and  can be resolved into two rectangular components.
Components of (i)  along PX
(ii)  along PY
Components of E_B
(i)  along PX
(ii)  along YP
Vertical components being equal and opposite cancel each other. Therefore, net electric field intensity along PX 
or,
If l << r so that it can be neglected, then
Q5: A dipole, with a dipole moment of magnitude p, is in stable equilibrium in an electrostatic field of magnitude E. Find the work done in rotating this dipole to its position of unstable equilibrium.
Sol:
For stable equilibrium, angle between p and E is 0º
For unstable equilibrium
Work done in rotating the dipole from angle θ₁ to θ₂

Q6: Deduce the expression for the electric field E due to a system of two charge q1 and q2 with 
position vectors r1 and r2 at a point ‘r’ with respect to common origin. 
Sol:

Electric field intensity at point P due to q₁,

Similarly,
Therefore, Net electric field intensity at point P,

Q7: Two charge –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x << d) perpendicular to the line joining the two fixed charged as shown in Fig. Show that q will perform simple harmonic oscillation and also find its time period.
Sol:

To show that q will perform simple harmonic oscillation and derive the expression for its time period, we can follow these steps: Let the force on charge q due to one of the fixed charges (−q) be F₁ and the force due to the other fixed charge be F₂. The net force on charge q is given by:

The force between two point charges is given by Coulomb’s Law:

Where k is the Coulomb’s constant, q₁ and q₂ are the magnitudes of the charges, and r is the separation between the charges. Let the distance of charge q from each of the fixed charges be d. So, the force F₁ and F₂ can be written as:
Since both forces are in opposite directions and have the same magnitude, the net force F_net on charge q will be zero. Now, when charge q is displaced slightly by x in the perpendicular direction (as shown in the figure), the net force F_net will not remain zero. There will be a restoring force that brings q back to the equilibrium position. The restoring force can be found using Hooke’s Law:
where k is the effective force constant. The effective force constant k can be determined by considering the net force on charge q when it is displaced by x. We can find this by using the concept of superposition:

Since F₁ and F₂ were calculated earlier, we can substitute their values:
Now, solving for k:

The angular frequency ω of the simple harmonic oscillation can be written as:

Substitute the value of K: 

The time period T of the oscillation is given by:

Substitute the value of ω: