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NEET PHYSICSELECTRIC CHARGES AND FIELDSEasy

Question

A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is:

A

3E along KO

B

E along OK

C

E along KO

D

3E along OK

Step-by-Step Solution

The electric field at the center of a uniformly charged circular ring is zero because the fields due to diametrically opposite elements cancel each other out. This implies that the electric field produced by any segment of the ring (AKB) must be equal in magnitude and opposite in direction to the electric field produced by the remaining part of the ring (ACDB) to result in a net zero field.

Mathematically, Etotal=EAKB+EACDB=0\vec{E}_{total} = \vec{E}_{AKB} + \vec{E}_{ACDB} = 0, which implies EACDB=EAKB\vec{E}_{ACDB} = -\vec{E}_{AKB}.

Given that the field due to AKB is EE. Since the charge is positive, the field direction due to segment AKB is directed away from the segment, i.e., from K towards center O (along KO). To cancel this, the field due to the rest of the ring (ACDB) must have the same magnitude EE but point in the opposite direction, i.e., from center O towards K (along OK).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTRIC CHARGES AND FIELDS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTRIC CHARGES AND FIELDSconductingradiuschargeelectriccentre

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