Question

Equal charges, q, are placed at the vertices of a hexagon from infinity. What is the work done to accomplish this?

please help!!

Answer 1

Work done is equal to the electric potential energy of system

For the following case :

Let us assume the edge length of hexagon as 'a' metres.

Let r₁₂ denote distance between charges at vertex 1 and 2

As we can see from the diagram,

r₁₂ = r₂₃ = r₃₄ = r₄₅ = r₅₆ = r₁₆ = a

r₁₃ = r₁₅ = r₂₄ = r₂₆ = r₃₅ = r₄₆ = a*3^1/2

r₁₄ = r₂₅ = r₃₆ = 2a

In this given system we have 6 charges, so the equation for total energy of system will contain 15 terms , because there are 15 distinct pairs of charges.

Those 15 terms get divided into 3 different groups, depending on the distance between charges

For the charges separated by distance 'a' , total energy = 6*k*q²/a

Here we multiply by 6 because we have 6 different pairs which are 'a' metres apart

Similarly,

For the charges separated by distance '2a' , total energy = 3*k*q²/2a

For the charges separated by distance '3^1/2a' , total energy = 6*k*q²/3^1/2a

Thus, total energy of system = kq²/a (6+ 3/2 + 6/3^1/2) = 10.96*k*q²/a Joules

Here, k = permitivitty of free space = 9*10⁹ , unit of q is Coulomb and a is metres