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`        circular wire loop of radius "a" 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.`
7 years ago

510 Points
```										total charge=Q
charge per unit length=Q/2pia
charge contained by dL lenght of element = dq=QdL/2pia ...........1
now
net electric field at the center of ring is E= 0 ....
when dL element is cut from the ring then let the field of remaining is ER..
so we can say that total electric field at center = ER + electric field due to dL element
(total electric field is vector sum of all components)
0=ER + kdq/a2
ER= -(KQdL/2pia3)

```
7 years ago
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