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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

For a uniformly charged ring of radius R , determine the distance...

For a uniformly charged ring of radius RR, determine the distance from its centre along the axis where the electric field achieves its maximum value.

Options

  1. A2R2R
  2. BR2\dfrac{R}{2}
  3. CR2R\sqrt{2}
  4. DR2\dfrac{R}{\sqrt{2}}

Correct answer

D. R2\dfrac{R}{\sqrt{2}}

Step-by-step solution

The electric field at a distance xx from the centre of a uniformly charged ring of radius RR and charge QQ on its axis is given by

E=kQx(R2+x2)3/2E = \dfrac{kQx}{(R^2 + x^2)^{3/2}}

For the electric field to be maximum, dEdx=0\dfrac{dE}{dx} = 0

ddx[x(R2+x2)3/2]=0\dfrac{d}{dx} \left[ \dfrac{x}{(R^2 + x^2)^{3/2}} \right] = 0

(R2+x2)3/2⋅1−x⋅32(R2+x2)1/2⋅2x(R2+x2)3=0\dfrac{(R^2 + x^2)^{3/2} \cdot 1 - x \cdot \dfrac{3}{2}(R^2 + x^2)^{1/2} \cdot 2x}{(R^2 + x^2)^3} = 0

(R2+x2)1/2[R2+x2−3x2]=0(R^2 + x^2)^{1/2} [R^2 + x^2 - 3x^2] = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2x = \dfrac{R}{\sqrt{2}}
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