Question:
A NONCONDUCTING RING OF RADIUS R HAS UNIFORMLY DISTRIBUTED VE CHARGE Q . A SMALL PART OF THE RING OF LENGTH d , IS REMOVED [DR] THE ELECTRIC FIELD AT THE CENTRE OF THE RING WILL NOW BE
Answer:
In case of a non-conducting ring of radius R, when charge is uniformly distributed, the charge at the centre of the ring will be zero.
Now when a small element dr is removed from the ring, the total field at the centre will be due to the non-conducting ring without the element dl. Let us suppose the radius of the ring is a, then the charge due to the element dr will be dq = q dl / 2 ∏ a -------------- Eqn (1)
The ring is composed of two parts: element dl + rest of the element in the ring. Hence, field at the centre = field due to element dl + field due to the rest of the element in the ring
Total field at centre due charged ring = Field due to element dl + Field due to the rest of the element in the ring say ER
Zero = k dq / a2 + ER
ER = -k dq / a2
= -k / a2 ( q dl / 2 ∏ a)
= -k q dl / 2 ∏ a3
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