Electric Field Due to a Ring of Charge

Description:

To find a net electric field at a point ‘P’ at a distance ‘x’ from a charged metallic ring, it is needed to find the vector sum of all the electric field components due to each source charge on the metallic ring.

To find the net electric field at point ‘P’, charge over the ring at ‘o’ and ‘P’ are joined as shown in the figure.

By Pythagoras theorem the distance OP is −

OP = √r² + x²

Where, ‘r’ is radius of ring

‘x’ is distance between the center of the ring to point ‘P’.

The ring has a linear charge density ‘λ’.

An infinitely small length (dl) of the ring is taken as shown in the figure above.

Calculating electric field due to dl at ‘P’.

Since, dl is very small so the charge on it would be same as that in a point charge.

i.e, dE = 1/4πε₀λ dl/r² + x²r̂ ........(1)

Here,  dE is the very small part of  electric field due to dl.

Similarly,

At point O’, infinitely small length ,diametrically opposite to point O is taken. This also make electric field at ‘P’ such that magnitude remains same but in the direction dE₂ ,as shown in the figure.

Note −

• The magnitude of electric field dE→₁ and dE→₂ are same but in different direction.

• As the magnitudes are same so we take |dE₁| = |dE₂| = |dE|

Now,

dE₁ and dE₂ are resolved into two components  such that;

• sin θ component of dE→₁ and dE→₂ cancels each other as they are equal in magnitude but opposite in nature.

• cos θ component of dE→₁ and dE→₂ is added up.

Similarly,

Electric field component dE sinθ due to each charge on the ring is cancelled out whereas the component dE cosθ is added up.

So, Resultant electric field = ∑ dEcosθ

So, Electric field due to total charge is −

∑ dE = 1/4πε₀λ dl/r² + x²x/(r² + x²)^1/2.r̂

(from figure cosθ= x/(r²+x²)^1/2)

For the net electric field integrating the equation.

we get: ∮ dE = ∮ 1/4πε₀λ dl/r² + x²x/(r² + x²)^1/2.r̂

⇒ E = 1/4πε₀λ x/(r² + x²)^3/2∮ dl

Total length of a ring 2πr

E = 1/4πε₀λ x/(r² + x²)^3/2

But Total charge(q) in the ring =2πr λ

electric field due to ring at a distance x is −

E = 1/4πε₀qx/(r² + x²)^3/2

Note − More the value of x, less will be the electric field.

Special Cases −

1. When x>>>r

Neglecting ‘r’ with respect to ‘x’ in equation (1) we get,

E = 1/4πε₀q/x²

In such case it behaves like a point charge.

2. When x<<

Neglecting ‘x’ with respect to ‘r’ in equation (1) we get,

E = 0

Note − Electric field at the centre of the ring is always Zero.