Question

A positive point charge (q 7.2 108 C) is surrounded by an equipotential surface A, which has a radius of rA 1.8 m. A positive test charge (q0 4.5 1011 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done as the test charge moves from surface A to surface B is WAB8.1 109 J. Find rB.

Answer 1

Thus the radius $r_{B}$ is 1.2 m

Explanation:

The potential at point A due to point charge at center .

V₁ = $\frac{q}{4\pi\epsilon_0r_A }$ =  7.2 x 10⁻⁸ x 9 x 10⁹ [$\frac{1}{r_A}$]

Similarly

The potential at B

V₂ = $\frac{q}{4\pi\epsilon_0r_B }$ = 7.2 x 10⁻⁸ x 9 x 10⁹ [ $\frac{1}{r_B}$ ]

The potential difference

V = V₂ - V₁ = 648  [ $\frac{1}{r_B}$ - $\frac{1}{r_A}$ ]

The work done

W = q₀ V = 4.5 x 10⁻¹¹ x 648 [$\frac{1}{r_B}$ - $\frac{1}{r_A}$ ]

But W = 8.1 x 10⁻⁹ J

Thus equation is

8.1 x 10⁻⁹ = 4.5 x 10⁻¹¹ x 648 [$\frac{1}{r_B}$ - $\frac{1}{r_A}$ ]

0.28 = [ $\frac{1}{r_B}$ - $\frac{1}{1.8}$ ]  or  $\frac{1}{r_B}$ = 0.28 + 0.55 = 0.83

or $r_{B}$ =  1.2 m