Question

two small pithballs are 3 cm apart in air a d carry charges of 3x10^-9 c and 12x10^-9 c respectively.A. calculate the force of attraction between them B. the force of balls are touch and then seperste by a distance of 3cm

Answer 1

use Coulomb's law, F=(kq1q2)/r^2
from this you'll get F as 1.08 ×10^-7 N

Answer 2

GIVEN ;

Two small pith balls are 3 cm apart in air a d carry charges of 3x10^-9 c and 12x10^-9 c respectively.A.,B. the force of balls are touched and then seperate by a distance of 3cm

TO FIND :

calculate the force of attraction between them

SOLUTION :

◆Pith balls are conductors, as soon as they touch each other charges flow and both charges become equal.

◆That is , here charges on both balls are, Q = Qa + Qb /2

= (3 + 12) /2 × 10^-9. = 7.5 ×10^-9

◆Force of attraction ,

F = K QaQb / r^2

K - constant in air , 9×10^9

Q is equal in balls

r distance between them.

◆On substitution,

F = 9×10^9 × 7.5×7.5 ×10^-18 /(3×3×10^-4)

= 506.25 /9 ×10^-5 N

◆F=5.625 ×10^-6 N.

ANSWER :

Force,

F=5.625 ×10^-6 N.