Two spheres carrying charges -6 - C and -9 - C- separated by a distance d- experiences a force of repulsion F- When a charge of -3 - C is given to both the sphere and kept at the same distance as before- the new force of repulsion isA- F-qB- F -3C- FD- 3 F

F α Q_1Q_2 F/F_1= 6 × 9/3 × 6⇒ F_1 = F/3 ...

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Byju's Answer

Standard XII

Physics

Charging by Friction

Two spheres c...

Question

Two spheres carrying charges $+ 6 μ C a n d + 9 μ C$ , separated by a distance d, experiences a force of repulsion F. When a charge of $- 3 μ C$ is given to both the sphere and kept at the same distance as before, the new force of repulsion is

$\frac{F}{3}$

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$F$

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$\frac{F}{9}$

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$3 F$

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Solution

The correct option is A
$\frac{F}{3}$

$F α Q_{1} Q_{2} \frac{F}{F_{1}} = \frac{6 \times 9}{3 \times 6} \Rightarrow F_{1} = \frac{F}{3}$

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Two spheres carrying charges +6μC and +9μC, separated by a distance d, experiences a force of repulsion F. When a charge of −3μC is given to both the sphere and kept at the same distance as before, the new force of repulsion is

The correct option is A F3 Fα Q1Q2FF1=6×93×6⇒F1=F3

Two spheres carrying charges $+ 6 μ C a n d + 9 μ C$ , separated by a distance d, experiences a force of repulsion F. When a charge of $- 3 μ C$ is given to both the sphere and kept at the same distance as before, the new force of repulsion is

The correct option is A
$\frac{F}{3}$

$F α Q_{1} Q_{2} \frac{F}{F_{1}} = \frac{6 \times 9}{3 \times 6} \Rightarrow F_{1} = \frac{F}{3}$