Question 12 Marks
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
View full question & answer→$\frac{\text{Charge}}{\text{Unit length}}=\frac{\text{Q}}{2\pi\text{a}}=\lambda;$ Charge of $\text{d}\ell=\frac{\text{Qd}\ell}{2\pi\text{a}}\text{C}$Initially the electric field was ‘0’ at the centre. Since the element ‘dℓ’ is removed so, net electric field must $\frac{\text{K}\times\text{q}}{\text{a}^2}$
Where q = charge of element $\text{d}\ell$
$\text{E}=\frac{\text{Kq}}{\text{a}^2}$
$=\frac{1}{4\pi\in_0}\times\frac{\text{Q}\text{d}\ell}{2\pi\text{a}}\times\frac{1}{\text{a}^2}$
$=\frac{\text{Qd}\ell}{8\pi^2\in_0\text{a}^3}$
Where q = charge of element $\text{d}\ell$
$\text{E}=\frac{\text{Kq}}{\text{a}^2}$
$=\frac{1}{4\pi\in_0}\times\frac{\text{Q}\text{d}\ell}{2\pi\text{a}}\times\frac{1}{\text{a}^2}$
$=\frac{\text{Qd}\ell}{8\pi^2\in_0\text{a}^3}$
$\text{d}=\text{ut}+\Big(\frac{1}{2}\Big)\text{at}^2$
$\text{F}=\text{qE},\ \text{F}=-\text{Kx}$
Since it is a regular hexagon. So, it forms an equipotential surface. Hence the charge at each point is equal. Hence the net entire field at the centre is Zero.
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