Question

Three charges 1 x 10-9 C, 2 x 10-9 C and 3 x 10-9 C are placed at the corners of an equilateral triangle of side 100 cm. Calculate the potential at O point equidistant from three corners of the triangle.

Answer 1

Answer:54√3V

Please refer to attachment.

Answer 2

Answer:

Given:-

First charge q₁ =  1 x 10⁻⁹ C

Second charge, q₂ = 2 x 10⁻⁹ C

Thrid charge, q₃ =  3 x 10⁻⁹ C

We know that, electric potential, V =  1/4πξ₀ × q/r

Now, the equidistant point will be the centroid of the triangle, then the value of r = $\frac{1}{\sqrt{3} }$

Now put the given values of charges q₁,q₂ and q₃ in the above equation, we get.

V =  1/4πξ₀ × q/r

V =  1/4πξ₀ × [q₁ + q₂ + q₃]/r

V =  9 ×  10⁹ × √3  × [q₁ + q₂ + q₃]

V =  9 ×  10⁹ × √3  × [1 x 10⁻⁹+ 2 x 10⁻⁹ + 3 x 10⁻⁹]

V =  9 ×  10⁹ × √3  × [1 x 10⁻⁹+ 2 x 10⁻⁹ + 3 x 10⁻⁹]

V =  93.6 V

∴The value of the electric potential is equal to 93.6 V.

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