Respuesta :
Thus the radius [tex]r_{B}[/tex] is 1.2 m
Explanation:
The potential at point A due to point charge at center .
V₁ = [tex]\frac{q}{4\pi\epsilon_0r_A }[/tex] = 7.2 x 10⁻⁸ x 9 x 10⁹ [[tex]\frac{1}{r_A}[/tex]]
Similarly
The potential at B
V₂ = [tex]\frac{q}{4\pi\epsilon_0r_B }[/tex] = 7.2 x 10⁻⁸ x 9 x 10⁹ [ [tex]\frac{1}{r_B}[/tex] ]
The potential difference
V = V₂ - V₁ = 648 [ [tex]\frac{1}{r_B}[/tex] - [tex]\frac{1}{r_A}[/tex] ]
The work done
W = q₀ V = 4.5 x 10⁻¹¹ x 648 [[tex]\frac{1}{r_B}[/tex] - [tex]\frac{1}{r_A}[/tex] ]
But W = 8.1 x 10⁻⁹ J
Thus equation is
8.1 x 10⁻⁹ = 4.5 x 10⁻¹¹ x 648 [[tex]\frac{1}{r_B}[/tex] - [tex]\frac{1}{r_A}[/tex] ]
0.28 = [ [tex]\frac{1}{r_B}[/tex] - [tex]\frac{1}{1.8}[/tex] ] or [tex]\frac{1}{r_B}[/tex] = 0.28 + 0.55 = 0.83
or [tex]r_{B}[/tex] = 1.2 m
The radius [tex]r_{B}[/tex] is 1.2 m.
Equipotential surface :
The potential difference between points A and B is given as,
[tex]V=V_{B}-V_{A}=kq(\frac{1}{r_{B}}-\frac{1}{r_{A}} )[/tex]
Where k is coulombs constant, value of [tex]k=9*10^{9} Nm^{2}/C[/tex]
- Substitute values in above equation.
[tex]V=9*10^{9} *7.2*10^{-8}(\frac{1}{r_{B}}-\frac{1}{1.8} )\\ \\V=648*(\frac{1}{r_{B}}-\frac{1}{1.8} )[/tex]
- Work done is given as,
[tex]W=q_{0}*V[/tex]
The work done as the test charge moves from surface A to surface B is[tex]8.1*10^{-9}J[/tex].
[tex]2.92*10^{-8} (\frac{1}{r_{B}}-\frac{1}{1.8} )=8.1*10^{-9} \\\\ (\frac{1}{r_{B}}-\frac{1}{1.8} )=0.277\\\\\frac{1}{r_{B}} =0.277+\frac{1}{1.8}=0.832\\ \\r_{B}=1/0.832=1.2m[/tex]
Find out more information about the equipotential surface here:
https://brainly.com/question/12074544
