Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
[tex]V =\frac{Kq}{r}[/tex]
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
[tex]V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V[/tex]
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
[tex]V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V[/tex]
Total potential due to this charges = 4500 V + 6750 V = 11250 V
