It's still zero!

A circular wire loop of radius RR carries a total charge qq distributed uniformly over its length. A small length x<<Rx<<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 qxαπβε0Rδ\dfrac { qx }{ \alpha { \pi }^{ \beta }{ \varepsilon }_{ 0 }{ R }^{ \delta } } , find α+β+δ\alpha +\beta +\delta .

Assume the medium to be vacuum.

Clarification: ε0{ \varepsilon }_{ 0 } is the permittivity of free space.

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