Answer:
![F'=(F)/(4)](https://img.qammunity.org/2020/formulas/physics/high-school/ofmhnwsh0sieekpq7ylsr0b6mey36tg12e.png)
1/4 as large
Step-by-step explanation:
Lets take
q₁ and q₂ are the charge on the particle and d is the distance between them.
We know that electrostatics between two particle given as
![F=K(q_1q_2)/(d^2)](https://img.qammunity.org/2020/formulas/physics/high-school/tp5hjbawkx3y64kyyd4gxum0hfw0ebw5yf.png)
When distance become double d'= 2 d
So new force F'
![F'=K(q_1q_2)/(d'^2)](https://img.qammunity.org/2020/formulas/physics/high-school/rnyjtnlpmxmhzcvbywgfszvq0q0ajs4azy.png)
![F'=K(q_1q_2)/((2d)^2)](https://img.qammunity.org/2020/formulas/physics/high-school/6s8tzgzugavxutr96icsj3we06ycklli95.png)
![F'=K(q_1q_2)/(4d^2)](https://img.qammunity.org/2020/formulas/physics/high-school/4jq438bubcgtalc4v8kohh2wymrby131h0.png)
![F'=(F)/(4)](https://img.qammunity.org/2020/formulas/physics/high-school/ofmhnwsh0sieekpq7ylsr0b6mey36tg12e.png)
It means that force will become one forth of initial force.
So answer is 1/4 as large .