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< 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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