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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. 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.
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.
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)
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)
Dear student,Please find the solution to your problem. Total charge=Qcharge per unit length=Q/2πacharge contained by dL lenght of element = dq = QdL/2πa ...........1now,net electric field at the center of ring is E= 0when dL element is cut from the ring then let the field of remaining is ERSo 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/a2ER = – (KQdL/2πa3) Thanks and regards,Kushagra
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