Three charges +q, +q and –q are kept at the vertices of an equilateral triangle of 10 cm side. The potential at the mid point in between +q and +q if q =5µC .

Complete Solution:

Given Data:

• Charges at vertices: +q, +q, -q
• Side length of the equilateral triangle: 10 cm
• Charge q = 5 µC

We need to find the electric potential at the midpoint between the two +q charges.

The midpoint lies on the line joining the two +q charges. Since this is an equilateral triangle, the midpoint is 5 cm from each +q charge.

The electric potential V at a point due to a point charge Q at a distance r is given by:

V = kQ/r

where k is the Coulomb constant, approximately 9 × 10⁹ N m²/C².

• Potential due to the first +q charge at a distance of 5 cm (0.05 m): V₁ = kq / 0.05
• Potential due to the second +q charge at a distance of 5 cm (0.05 m): V₂ = kq / 0.05

Since both charges are +q and are at equal distances from the midpoint, the potentials due to these charges add up:

Potential due to the third charge -q (at the opposite vertex of the triangle)

The distance from this -q charge to the midpoint is the height of the equilateral triangle:

Height = (√3 / 2) × side = (√3 / 2) × 0.1 m = 0.0866 m

The potential due to the -q charge at this distance:

V₃ = -kq / 0.0866

Using k = 9 × 10⁹ N m²/C² and q = 5 × 10^-6 C:

• Calculate V₁ + V₂: V₁ + V₂ = 2 × (9 × 10⁹ × 5 × 10^-6 / 0.05) = 1.8 × 10⁶ V
• Calculate V₃: V₃ = (9 × 10⁹ × -5 × 10^-6) / 0.0866 = -5.2 × 10⁵ V

Total Potential at the Midpoint

V_total = V₁ + V₂ + V₃ = 1.8 × 10⁶ - 5.2 × 10⁵ = 1.28 × 10⁶ V

Final Answer:

The potential at the midpoint between the two +q charges is approximately 1.28 × 10⁶ V.