Question

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

Answer 1

Answer:

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