Login

the sum of two point charges is 6 micro coloumb.they attract each other with a force of 0.9N,when kept 40cm apart in vacuum.calculate the charges.please do not give links​

Dear student
as there is force of attraction so one of the charge is positive and the other is negative.Let one charge is q1 and the other one is-q2.Let( q1-q2)=6μC..1let q2=q so q1=6+qF=14π∈q×(6+q)×10-120.4×0.40.9=9×109q×(6+q)×10-120.4×0.49×16×10-3=9×10-3×q×(6+q)q×(6+q)=16q2+6q-16=0q2+8q-2q-16=0(q+8)×(q-2)=0q=2 or q=-8neglecting the negative oneq2=2sign wise q2=-2q1=8Regards
 

  • 3
View Full Answer
let the two charges be q1 and q2. given q1+q2 = 6 micro C, then q1 = 6 micro C - q2
​F = 0.9 N ; distance r = 0.4m

​by coloumb's law, F = (kq1q2)/(r xr)
​                              0.9 = (9 x 10 power 9 x  (6 x 10 power -6  - q2) x q2) / 0.4 0.4
​                             0.36 =54000q2 - (q2x q2)
                             (q2 x q2) - 54000q2 + 0.36 = 0 (quadratic eq)
​                       after using quadratic eq formula,
                                     q2 =  54000 ( + or -) 53999/ 2
                                           in + case q2 = 54000 + 53999/ 2 = 53999 (but point charge cannot have such a huge value)
​                                           in - case q2 = 54000 - 53999/2 = 1/2 ( which is possible)
​                                           therefore, q2 = 0.5 C
                                                            q1 = (6 x 10 power -6) - 0.5 = -0.49 C
    
  • 0
Regards

  • 2
What are you looking for?