2 equal point charges of 20nc are located at x=2,3m respectively what is the potential at origin?

Question

Answer

Calculation of Potential at the Origin due to Two Point Charges

Given Parameters:

Two point charges of 20nC are located at x = 2m and x = 3m respectively.

The distance between the charges is 1m.

We need to find the potential at the origin.

Concept of Potential:

Potential is a scalar quantity that determines the amount of work needed to bring a unit positive charge from infinity to a point in an electric field without any acceleration.

The potential due to a point charge q at a distance r from it is given by:

V = kq/r, where k is the Coulomb constant (9 x 10^9 Nm^2/C^2).

The potential due to multiple point charges is the algebraic sum of the potentials due to individual point charges.

Calculation:

Let's assume that the charge at x = 2m is q1 and the charge at x = 3m is q2.

The distance between the charges is 1m.

The potential at the origin due to q1 is:

V1 = kq1/r1, where r1 = 2m (distance between q1 and the origin).

V1 = (9 x 10^9 Nm^2/C^2) x (20 x 10^-9 C) / 2m.

V1 = 90 V.

The potential at the origin due to q2 is:

V2 = kq2/r2, where r2 = 3m (distance between q2 and the origin).

V2 = (9 x 10^9 Nm^2/C^2) x (20 x 10^-9 C) / 3m.

V2 = 60 V.

The net potential at the origin due to both charges is:

V = V1 + V2.

V = 90 V + 60 V.

V = 150 V.

Explanation:

Electric potential due to a point charge is the work done to bring a unit positive charge from infinity to a point in an electric field without any acceleration.

For two or more point charges, the total potential is the algebraic sum of the potentials due to individual point charges.

In this case, the potential at the origin due to both charges is 150 V.

Answer

But options are 270v 200v 261v 0v

Answer

261v

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