He absolute potential at a distance of 2.0 m from a positive point charge is 100 V. What is the absolute potential 4.0 m away from the same point charge?

50 V

Explanation:

The formula electric potential is given as,

V = kq/r Equation 1

q = Vr/k Equation 2

Where q = charge at that point, V = Electric potential, k = coulombs constant, r = distant.

Given: V = 100 V, r = 2.0 m, k = 9.0×10⁹ Nm/C².

Substitute into equation 1

q = (100×2)/(9.0×10⁹ )

q = (200/9)(10⁹)

q = 22.22×10⁻⁹

q = 2.22×10⁻⁸ C.

The potential at point 4.0 m

Given: r = 4.0 m, q = 2.22×10⁻⁸ C, k = 9.0×10⁹ Nm²/C²

Substitute into equation 2

V = 9.0×10⁹(2.22×10⁻⁸)/4

V = 49.95 V

V ≈ 50 V

Hence the potential = 50 V

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000