Question

Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each charge will be

Answer 1

Answer:

The force on each charge = 16 N

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

When d = 1 m, F = 1 N,

Constant: 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 2,

q₁q₂ = 1×1²/9×10⁹

q₁q₂ = 1/9×10⁹  C²

When d = 1/4 m, q₁q₂ = 1/9×10⁹  C² and 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 1

F =  9×10⁹×1/9×10⁹ /(1/4)²

F = 1/(1/16)

F = 16 N

Therefore the force on each charge = 16 N