Answer:
(a) V1 = 8990.00 V
V2 = 8960.13 V
Explanation:
Parameters given:
q =3 mC
k = 8.99 * 10⁹ Nm²/C²
x1 = 3 m
x2 = 3.01 m
Electric potential is given as:
V = kq/r
Where
k = Coulombs constant
q = charge
r = distance
Potential at x1 is:
V1 = (8.99 * 10⁹ * 0.000003)/(3)
V1 = 8990.00V
Potential at x2 is:
V2 = (8.99 * 10⁹ * 0.003)/(3.01)
V2 = 8960.13 V
Answer:
a) 8,990.00 V b) 8,960.13 V
Explanation:
a) The potential due to a point charge, can be found from the expression of Coulomb's Law, as follows:
[tex]V = \frac{k*q}{r}[/tex]
where k = 8.99*10⁹ N*m²/C², q = 3.00*10⁻⁶ C, and r = 3.00 m.
Replacing by this values, we can find the potential V as follows:
[tex]V = \frac{8.99e9 N*m2/C2*3.00e-6C}{3.00m} = 8,990.00 V[/tex]
b) Repeating the process for r = 3.01m:
[tex]V = \frac{8.99e9 N*m2/C2*3.00e-6C}{3.01m} = 8,960.13 V[/tex]
