Question

two charges 1 uC and 5 uC separated by 20 cm the ration of electrical forces acting on them will be​

Answer 1

GiveN :

• q1 = 1μC = 10^-6 C
• q2 = 5μC = 5*10^-6 C
• Distance (r) = 20cm = 0.2 m

To FinD :

• Magnitude of electrostatic force between them

SolutioN :

⇒F12 = 1/4πϵo * q1q2/r²

⇒F12 = 9*10^9 * (10^-6 × 5*10^-6)/(0.2)²....(1)

Similar value will be in F21

⇒F21 = 9 * 10^9 * ( 5*10^-6 * 10^6)/(0.2)² ...(2)

__________________

Simply divide (1) and (2)

⇒F12/F21 = 1:1

As values of F12 and F21 will be same

So, ratio of forces is 1:1