Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is

|F|=K|QQ′|d2,

where K=1/4πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -10.0 nC , is located at x1 = -1.695 m ; the second charge, q2 = 30.5 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.245 m ?

Your answer may be positive or negative, depending on the direction of the force.

The net force exerted by the two charges is 10.97 x 10⁻⁵ N along negative x-direction.

Explanation:

K=1/4πϵ0, where ϵ0=8.854×10−12C2

K = 9x10⁹

The electric force on point charge q₃ due to charge q₂ is

F₃₂ = kq₃q₂ / (1.245)²

     = (9x10⁹ * 49.5x10⁻⁹ * 30.5x10⁻⁹) / (1.245)²

     = 13,587.75 x 10⁻⁹ / 1.55

     = 8.76629 x 10⁻⁵ N

The electric force on point charge q₃ due to charge q₁ is

F₃₁ = kq₃q₁ / (1.695 - 1.245)²

     = (9x10⁹ * 49.5x10⁻⁹ * 10.0x10⁻⁹) / 0.2025

     = 2.2000 x 10⁻⁵ N

The net electric force on point charge q₃ is

F₃ = -F₃₁ - F₃₂

    = - 8.76629 x 10⁻⁵ N - 2.2000 x 10⁻⁵ N

    = 10.97 x 10⁻⁵ N along negative x-direction

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

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