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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.
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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