Question

A circular wire loop of radius a carries a total charge Q distributed uniformly over its length. A small length DL of the wire is cut off. Find the electric field at the centre due to the remaining wire.

Answer 1

Step 1: Given that:

Radius of the circular wire loop= $a$

Charge on the wire= $Q$

Length of small element cut off from the wire= $dL$

Step 2: Formula used:

The electric field at the centre of a ​circular loop is zero.

The total wire now consists of two parts after cutting; the cut off part of the length $dL$ and the remaining part.

Thus,

The total electric field at the centre due to the whole circular wire loop($E$) = $0$

The electric field due to the small length element($E_{lengthelement}$ ) +The electric field due to the remaining part of the wire($E_{remainingwire}$ )=$0$

The electric field due to remaining part of the wire= - the electric field due to the length element $dL$

That is

$E_{remainingpart}=-E_{lengthelement}$............1)

$Or,$

$\left|E_{remainingpart}\right|=\left|E_{lengthelement}\right|$...............2)

Step 3: Calculation of electric field at the centre due to the remaining wire:

Electric field of small element $dL$ will be calculated as follows:

The charge density of the circular wiore loop($\lambda$)= $\frac{Chargeontheloop}{Totallengthofthewire}=\frac{Chargeontheloop}{Circumferenceofthecircularloop}$

$\lambda =\frac{Q}{2\pi a}$

Therefore the charge on the small length element of the loop will be:

$dQ=\lambda dL$

$dQ=\frac{QdL}{2\pi a}$

The small length element can be considered as a point charge.

Thus,

The electric field due to the point charge at the centre of the loop will be:

$E_{lengthelement}=\frac{1}{4\pi {\epsilon }_0}\frac{dQ}{a^2}$

$E_{lengthelement}=\frac{1}{4\pi {\epsilon }_0}\times \frac{QdL}{2\pi a\times a^2}$

$E_{lengthelement}=\frac{QdL}{8{\pi }^2{\epsilon }_0{a}^3}$

Therefore, from equation 1) and 2)

The electric field due to the remaining wire loop will be:

$E_{remainingwire}=\frac{QdL}{8{\pi }^2{\epsilon }_0{a}^3}$

$(DirectedoppositetothedirectionofelectricfieldduetothesmalllengthelementdL.)$