# 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.

509 Points
13 years ago

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)

4 years ago
Dear student,

Total charge=Q
charge per unit length=Q/2πa
charge contained by dL lenght of element = dq = QdL/2πa ...........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/2πa3)

Thanks and regards,
Kushagra