Physics, asked by guyvishal3457, 1 year ago

Two positive charges separated by a distance 2m each other with a force of 0.36N. If the combined charge is 26 microcoloumbs , the charges are?

Answers

Answered by CBSEMP
26
Let the two charges be Q and q
  Force of repulsion between these charges is 0.36 N
 14πε0×Qq(2m)2=0.36N9×109×Qq4m2=0.36N

Qq=0.36×49×109=16×10-11 C2  ---(1 )

Given that sum of charges is 26×10-6C       

Q+q=26×10-6C  ,   Q=26×10-6-q

Putting value of Q in equation (1)    

  q(26×10-6-q)=16×10-11     

 q2-26×10-6q+16×10-11=0       

q2 -16*10-6-10*10-6 q+16*10-11=0

q(q- 16* 10-6)-10 *10-6(q-16*10-6 )=0

(q-16*10-6)(q-10*10-6)=0

 q=16×10-6C                                    or        q=10×10-6C

If q=16*10-6C  , Q=10*10-6C

If q=10*10-6C   , Q=16*10-6C

Value of two charges are 16×10-6C and 10×10-6C


Answered by saibhavay
34

HOPE it will help you by this answer

Attachments:
Similar questions