A positive charge Q1 = +4.80 nC is located at x1 = -2.00 m, a negative charge Q2 = -6.30 nC is located at x2 = 3.00 m, and a positive charge Q3 = 5.40 nC is located at x3 = 9.00 m. Find the net electric potential at x = 0.
Question
A positive charge Q1 = +4.80 nC is located at x1 = -2.00 m, a negative charge Q2 = -6.30 nC is located at x2 = 3.00 m, and a positive charge Q3 = 5.40 nC is located at x3 = 9.00 m. Find the net electric potential at x = 0.
Solution
The electric potential V due to a point charge Q at a distance r is given by the equation:
V = kQ/r
where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2).
We can use this equation to find the electric potential at x = 0 due to each of the three charges, and then add these potentials together to find the net electric potential.
- For Q1 = +4.80 nC located at x1 = -2.00 m, the distance r1 from x = 0 is 2.00 m. The electric potential V1 is:
V1 = kQ1/r1 = (8.99 x 10^9 N m^2/C^2)(4.80 x 10^-9 C)/(2.00 m) = 21.57 V
- For Q2 = -6.30 nC located at x2 = 3.00 m, the distance r2 from x = 0 is 3.00 m. The electric potential V2 is:
V2 = kQ2/r2 = (8.99 x 10^9 N m^2/C^2)(-6.30 x 10^-9 C)/(3.00 m) = -18.97 V
- For Q3 = +5.40 nC located at x3 = 9.00 m, the distance r3 from x = 0 is 9.00 m. The electric potential V3 is:
V3 = kQ3/r3 = (8.99 x 10^9 N m^2/C^2)(5.40 x 10^-9 C)/(9.00 m) = 5.39 V
The net electric potential at x = 0 is the sum of these three potentials:
V_net = V1 + V2 + V3 = 21.57 V - 18.97 V + 5.39 V = 7.99 V
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