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.

Explanation:

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

V₁ = =  7.2 x 10⁻⁸ x 9 x 10⁹ []

Similarly

The potential at B

V₂ = = 7.2 x 10⁻⁸ x 9 x 10⁹ [ ]

The potential difference

V = V₂ - V₁ = 648  [ - ]

The work done

W = q₀ V = 4.5 x 10⁻¹¹ x 648 [ - ]

But W = 8.1 x 10⁻⁹ J

Thus equation is

8.1 x 10⁻⁹ = 4.5 x 10⁻¹¹ x 648 [ - ]

0.28 = [ - ]  or   = 0.28 + 0.55 = 0.83

or =  1.2 m

Distance traveled is the average velocity times time.  When something is dropped (or thrown), it accelerates downward in accord with the acceleration of gravity ( g = 9.8m/sec2)

The increase in velocity while falling is simply t * 9.8,  t = 2, so the increase is 19.6 m/s.  The initial velocity is 3, so the final is 22.6, and the average is 25.6/2 = 12.8m/s

Therefore, d = v(average) * t = 12.8 * 2 = 25.6m

The equivalent formula is:

d = v(initial)*t + (a*t2) / 2

d = 3*2 + 4.9 * 4 = 25.6

Explanation:

Distance traveled is the average velocity times time.  When something is dropped (or thrown), it accelerates downward in accord with the acceleration of gravity ( g = 9.8m/sec2)

The increase in velocity while falling is simply t * 9.8,  t = 2, so the increase is 19.6 m/s.  The initial velocity is 3, so the final is 22.6, and the average is 25.6/2 = 12.8m/s

Therefore, d = v(average) * t = 12.8 * 2 = 25.6m

The equivalent formula is:

d = v(initial)*t + (a*t2) / 2

d = 3*2 + 4.9 * 4 = 25.6