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Igoryamba
3 years ago
9

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

se charges at a point at on the x-axis?
Physics
1 answer:
Airida [17]3 years ago
6 0

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;

V =\frac{Kq}{r}

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:

V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V

Total potential due to this charges  = 4500 V + 6750 V = 11250 V

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