Three point charges Q+10x10-12C, Q₂= -6 x 10-12C, and Q3+10x10-12C, are situated at the corners of an equilateral triangle of side 10 m in vacuum. The electric potential at the centre of the triangle is (a)
10 x 10-3v (b) 5 x 10-3v (C) 6.5 x 10-3v (d) 7.8 x10-3v
10 V
To find the electric potential at the center of the triangle, we need to calculate the electric potential due to each individual point charge and then sum them up.
The electric potential V due to a point charge Q at a distance r is given by V = kQ/r, where k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2).
Let's denote the distance from each charge to the center of the triangle as R. Since the triangle is equilateral, R = side length / (√3) = 10 m / √3.
Calculating the electric potential due to each point charge:
V₁ = k(Q₁) / R = (8.99 x 10^9)(10 x 10^-12) / (10 / √3) = 8.99 x 10^-3 V
V₂ = k(Q₂) / R = (8.99 x 10^9)(-6 x 10^-12) / (10 / √3) = -5.39 x 10^-3 V
V₃ = k(Q₃) / R = (8.99 x 10^9)(10 x 10^-12) / (10 / √3) = 8.99 x 10^-3 V
The total electric potential at the center of the triangle is the sum of the potentials due to each point charge:
V_total = V₁ + V₂ + V₃ = 8.99 x 10^-3 V - 5.39 x 10^-3 V + 8.99 x 10^-3 V = 12.59 x 10^-3 V
Therefore, the electric potential at the center of the equilateral triangle is 12.59 x 10^-3 V, which is approximately equal to 12.6 x 10^-3 V.
The closest answer choice is (d) 7.8 x 10^-3 V.