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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

Two spheres of radius aa and bb respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is:

A

a/b

B

b/a

C

a²/b²

D

b²/a²

Step-by-Step Solution

When two charged conductors are connected by a wire, charge flows between them until they acquire the same electric potential (V1=V2V_1 = V_2) [Source 53].

  1. Potential Condition: For spheres of radius aa and bb with charges QaQ_a and QbQ_b: Va=Vb⇒kQaa=kQbb⇒QaQb=abV_a = V_b \Rightarrow \frac{kQ_a}{a} = \frac{kQ_b}{b} \Rightarrow \frac{Q_a}{Q_b} = \frac{a}{b} [Source 41].

  2. Electric Field Ratio: The electric field at the surface of a conducting sphere is given by E=kQR2E = \frac{kQ}{R^2} [Source 41]. The ratio of the fields is: EaEb=kQa/a2kQb/b2=(QaQb)×(b2a2)\frac{E_a}{E_b} = \frac{kQ_a / a^2}{kQ_b / b^2} = \left( \frac{Q_a}{Q_b} \right) \times \left( \frac{b^2}{a^2} \right)

  3. Substitution: Substituting the charge ratio derived in step 1: EaEb=(ab)×(b2a2)=ba\frac{E_a}{E_b} = \left( \frac{a}{b} \right) \times \left( \frac{b^2}{a^2} \right) = \frac{b}{a}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. 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.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEspheresradiusrespectivelychargedjoined

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