Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

50V

Explanation:

The absolute potential, (or electric potential), V, of a point charge Q from a distance r is given by;

V = $\frac{k Q}{r}$   ---------------------(i)

Where;

k = electric constant and has a value = 9.0 x 10⁹Nm²/C².

From the question,

V = 100V when r = 2.0m

But the quantity of the charge Q has not been given.

Lets calculate that first.

Substitute the values of V, r and k into equation (i) as follows;

=> 100 = $\frac{9 * 10^{9} * Q }{2.0}$

=> Q = $\frac{100 * 2.0}{9*10^{9} }$

=> Q = 22.22 x $10^{-9}$ C

The quantity of charge Q is 22.22 x $10^{-9}$ C

Now, to get the absolute potential (V) 4.0m away from the same point charge Q,

We substitute Q = 22.22 x $10^{-9}$ C, r = 4.0m and k = 9.0 x 10⁹Nm²/C² into equation(i) as follows;

=> V = 9.0 x 10⁹ x $\frac{22.22*10^{-9} }{4.0}$

=> V =  49.995V

=> V = 50V (to the nearest whole number)

Therefore the absolute potential (V) 4.0m away from the same point charge is 50V.