Question

The force between two points charge in air is 100N calculate the force if the distance between them become s double

Answer 1

Answer:

Explanation: we know F = kq²/r² means i.e inversely relate to r²

            hence F' = F/4 = 25 N

  hope u get it mate........................

Answer 2

Given :

• Force between two point charges is 100 N

To Find :

• Force between charges if distance between them is doubled.

Solution :

As we know that,

$\implies \sf{F \: = \: \dfrac{1}{4 \pi \epsilon _o} \dfrac{q_1 q_2}{r^2} \: \: \: \: \: \: ...(1)}$

Here,

• F is force
• r is distance between the charges
• And q1, q2 are charges

_________________________________

If Distance is Doubled then force is :

$\implies \sf{F' \: = \: \dfrac{1}{4 \pi \epsilon_o} \dfrac{q_1 q_2}{(r')^2}} \\ \\ \implies \sf{F' \: = \: \dfrac{1}{4 \pi \epsilon_o} \dfrac{q_1 q_2}{(2r) ^2}} \\ \\ \implies \sf{F' \: = \: \dfrac{1}{4 \pi \epsilon_o} \dfrac{ q_1 q_2}{4 r^2}} \\ \\ \implies \sf{F' \: = \: \dfrac{F}{4}} \\ \\ \implies \sf{F' \: = \: \dfrac{100}{4}} \\ \\ \implies \sf{F' \: = \: 25 \: N}$