Fe=K Q1/Q2/d2 Q1 is the first charge Q2 is the second charge d is the distance K= 9x10^9 NM^2/C2 Now let’s plug the numbers Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment And we can curtailment the M^2and the other M^2 Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3 Let’s multiply the (9)(2)(8)=144 And add the exponents (9)+(-4)+(-4)=1 So now we got 144x10N divide by the distance which is 0.3 144x10N / 0.3 = 4800N Hope it helps u understand :)
