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Chapter 11: Problem 1547

Two uniformaly charged spherical conductors \(\mathrm{A}\) and \(\mathrm{B}\) having radius \(1 \mathrm{~mm}\) and \(2 \mathrm{~mm}\) are separated by a distance of \(5 \mathrm{~cm} .\) If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres \(\mathrm{A}\) and \(\mathrm{B}\) is..... (A) \(4: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(1: 4\)

Short Answer

Expert verified
The ratio of the magnitude of the electric fields at the surfaces of spheres A and B is \(2:1\).

Step by step solution

01

Understanding the equilibrium condition

In equilibrium condition, the potential of conductors connected by a conductor remains the same. For spheres A and B, this means that the potential on A is equal to the potential on B. Therefore, we have: \(ER_{A} = ER_{B}\) or \(E_{A} = \frac{R_{B}}{R_{A}}E_{B}\) (1)
02

Calculate the ratio of the electric fields

The ratio of the magnitude of the electric fields at the surfaces of spheres A and B can be calculated by substituting the radius of sphere A \(R_{A}=1mm\) and B \(R_{B}=2mm\) into equation (1). The result is \(E_{A}/E_{B} = \frac{2mm}{1mm} = 2\). The correct choice is (C) \(2: 1\). So, the ratio of the electric fields at the surfaces of the two spheres (A and B) is \(2:1\). This suggests that the electric field at the surface of sphere A is twice as strong as that at the surface of sphere B. This is due to the smaller radius of sphere A, which results in a stronger electric field.

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Most popular questions from this chapter

An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in an uniform electric field \(E=100 \mathrm{Vm}^{-1}\). A particle of mass \(1 \mathrm{~kg}\) and charge \(0.01 \mathrm{c}\) is allowed to slide down from rest from a height of \(1 \mathrm{~m} .\) If the coefficient of friction is \(0.2\) the time taken by the particle to reach the bottom is $\ldots \ldots . .$ sec (A) \(2.337\) (B) \(4.337\) (C) 5 (D) \(1.337\)

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