# It's still zero!

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

If the answer is in the form of $$\dfrac { qx }{ \alpha { \pi }^{ \beta }{ \varepsilon }_{ 0 }{ R }^{ \delta } }$$, find $$\alpha +\beta +\delta$$.

Assume the medium to be vacuum.

Clarification: $${ \varepsilon }_{ 0 }$$ is the permittivity of free space.

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