Question

two identical spheres charged with 200 microcoloumb and -100 microcoloumb are kept at a distance. The force acting on them is f1. They are connected using a conductor and then the conductor is removed. The new force f2 acting on them will be ​

Answer 1

Answer:

here,

F1 = 0.108 N

r = 0.5 m

F2 = 0.036 N

let charges on the spheres be q1 and q2 respectively.

charges after connection gets redistributed in the spheres , so let charge on each sphere be q.

from conservation of charge

q1 + q2 = 2q...........eq.1

again

F1 = k q1q2/r²...........eq.2

F2 = k q²/r².............eq.3

=> 0.036 = 9×10⁹ ×q²/(0.5)²

=> q² = 0.036 × (0.5)²/9×10⁹

=> q² = 1×10⁻¹²

=> q=1×10⁻⁶ C.............eq.4

from eq 2 and 3

F1/F2 = q1q2/q²

=> 3 = q1q2/q²

3q² = q1q2

=> q1q2 = 3q²

again,

(q1 ₋ q2)² = (q1 + q2)² ₋ 4q1q2

= (2q)² ₋ 3q²/r²

= q²/r²

by eq.4

(q1₋q2) = q/r

=1×10⁻⁶/0.5

=2×10⁻⁶..............eq.5

from eq.1 and eq.4

q1 + q2 = 4×10⁻⁶..............eq.6

adding 5 and 6

q1 = 3×10⁻⁶

q1 = 3μC

subtracting 5 from 6

q2 = 1×10⁻⁶

q2 = 1μC