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:
50V
Explanation:
The absolute potential, (or electric potential), V, of a point charge Q from a distance r is given by;
V = [tex]\frac{k Q}{r}[/tex] ---------------------(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 = [tex]\frac{9 * 10^{9} * Q }{2.0}[/tex]
=> Q = [tex]\frac{100 * 2.0}{9*10^{9} }[/tex]
=> Q = 22.22 x [tex]10^{-9}[/tex] C
The quantity of charge Q is 22.22 x [tex]10^{-9}[/tex] C
Now, to get the absolute potential (V) 4.0m away from the same point charge Q,
We substitute Q = 22.22 x [tex]10^{-9}[/tex] C, r = 4.0m and k = 9.0 x 10⁹Nm²/C² into equation(i) as follows;
=> V = 9.0 x 10⁹ x [tex]\frac{22.22*10^{-9} }{4.0}[/tex]
=> 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.
