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. *1 point- 48.7 V+ 48.7 V- 35.1 V+38.7 V
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. *1 point- 48.7 V+ 48.7 V- 35.1 V+38.7 V
Solution 1
The electric potential V at a point due to a point charge Q is given by the equation:
V = kQ/r
where:
- k is Coulomb's constant (8.99 x 10^9 N.m^2/C^2),
- Q is the charge,
- r is the distance from the charge to the point.
We can calculate the electric potential at x = 0 due to each charge, and then add them up to get the net electric potential.
-
For Q1 = +4.80 nC located at x1 = -2.00 m: V1 = kQ1/r1 = (8.99 x 10^9 N.m^2/C^2) * (4.80 x 10^-9 C) / 2.00 m = 21.6 V
-
For Q2 = -6.30 nC located at x2 = 3.00 m: V2 = kQ2/r2 = (8.99 x 10^9 N.m^2/C^2) * (-6.30 x 10^-9 C) / 3.00 m = -18.9 V
-
For Q3 = 5.40 nC located at x3 = 9.00 m: V3 = kQ3/r3 = (8.99 x 10^9 N.m^2/C^2) * (5.40 x 10^-9 C) / 9.00 m = 5.4 V
The net electric potential at x = 0 is the sum of these three potentials:
V_net = V1 + V2 + V3 = 21.6 V - 18.9 V + 5.4 V = 8.1 V
None of the provided options match this result. Please check the question and the provided options.
Solution 2
The electric potential V at a point due to a point charge Q is given by the equation:
V = kQ/r
where:
- k is Coulomb's constant (8.99 x 10^9 N.m^2/C^2),
- Q is the charge,
- r is the distance from the charge to the point.
We can calculate the electric potential at x = 0 due to each charge, and then add them up to get the net electric potential.
-
For Q1 = +4.80 nC located at x1 = -2.00 m: V1 = kQ1/r1 = (8.99 x 10^9 N.m^2/C^2) * (4.80 x 10^-9 C) / 2.00 m = 21.6 V
-
For Q2 = -6.30 nC located at x2 = 3.00 m: V2 = kQ2/r2 = (8.99 x 10^9 N.m^2/C^2) * (-6.30 x 10^-9 C) / 3.00 m = -18.9 V
-
For Q3 = 5.40 nC located at x3 = 9.00 m: V3 = kQ3/r3 = (8.99 x 10^9 N.m^2/C^2) * (5.40 x 10^-9 C) / 9.00 m = 5.4 V
The net electric potential at x = 0 is the sum of these three potentials:
V_net = V1 + V2 + V3 = 21.6 V - 18.9 V + 5.4 V = 8.1 V
None of the provided options match this result. Please check the question and the provided options.
Similar Questions
A positive charge Q1 = +4.08 nC is located at x1 = -2.00 m, a negative charge Q2 = -6.50 nC is located at x2 = 3.00 m, and a positive charge Q3 = 4.30 nC is located at x3 = 9.00 m. Find the net potential energy in the field.*1 point-84.25 nJ+84.25 nJ126. 67 nJ105.46 nJ
Two charges are set on the x-axis 11.4 cm away from each other. The charges are -6.10 nC and 21.4 nC. Calculate the electric potential at the point on the x-axis where the electric field due to these two charges is zero.
Four equal point charges Q =10 nC are place at the four corners of a square 2 m on a side. Find the electric potential at the centre of the square.*1 point25.53 V20.50 V35.23 V28.42 V
The figure below shows two point charges, 𝑄1 = +7.50 nC and 𝑄2 = −8.30 nC, separated bya distance of 5.00 mm.Calculate(a) the force acted upon 𝑄1.(b) the electric field strength at the centre between the point charges.(c) the electric potential at the centre between the point charges.
Two small spheres with charges q1 and q2 are 20 cm apart. A third sphere with a charge of −2.0 nC is placed directly between them.If the sphere with charge q2 experiences a zero net electric force, what is the value of q1?A.−8 nCB.4 nCC.8 nCD.16 nC
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.