Ans. 3.8 X IUL2. Two identical conducting spheres haviropposite sign a

1.	Ans. 3.8 X IUL2. Two identical conducting spheres haviropposite sign attract each other with a forcewhen separated by 0.5 m The spheres arenected by a conducting wire, which is thenand thereafter repel each otherwith a force of 0.036 N. Whatwere the initial charges on thespheres ?pheres having charges ofnith a force of 0.108he spheres are con-hich is then removedĂ2 cm1 cm
Answer» Herehere,F1 = 0.108 Nr = 0.5 mF2 = 0.036 Nlet charges on the spheres be q1 and q2 respectively.charges after connection gets redistributed in the spheres , so letcharge on each sphere be q.from conservation of chargeq1 + q2 = 2q...........eq.1 againF1 = k q1q2/r²...........eq.2F2 = 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 3F1/F2 = q1q2/q²=> 3 = q1q2/q²3q² = 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.4q1 + q2 = 4×10⁻⁶..............eq.6 adding 5 and 6q1 = 3×10⁻⁶q1 = 3μC subtracting 5 from 6q2 = 1×10⁻⁶q2 = 1μC

Ans. 3.8 X IUL2. Two identical conducting spheres haviropposite sign attract each other with a forcewhen separated by 0.5 m The spheres arenected by a conducting wire, which is thenand thereafter repel each otherwith a force of 0.036 N. Whatwere the initial charges on thespheres ?pheres having charges ofnith a force of 0.108he spheres are con-hich is then removedĂ2 cm1 cm

Answer»

Herehere,F1 = 0.108 Nr = 0.5 mF2 = 0.036 Nlet charges on the spheres be q1 and q2 respectively.charges after connection gets redistributed in the spheres , so letcharge on each sphere be q.from conservation of chargeq1 + q2 = 2q...........eq.1

againF1 = k q1q2/r²...........eq.2F2 = 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 3F1/F2 = q1q2/q²=> 3 = q1q2/q²3q² = 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.4q1 + q2 = 4×10⁻⁶..............eq.6

adding 5 and 6q1 = 3×10⁻⁶q1 = 3μC

subtracting 5 from 6q2 = 1×10⁻⁶q2 = 1μC

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