# Two identical balls carrying charges +6μC and +9μC separated by a distance d experiences a force of repulsion F. When a charge of -3μC is given to bot

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Two identical balls 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 balls and kept at the same distance as before, the new force of repulsion will be:

1. 3F
2. F/9
3. F
4. F/3

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Correct Answer - Option 4 : F/3

CONCEPT:

Coulomb's law:

• According to this law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between them.
• It acts along the line joining the two charges.
• Mathematically it is written as

$⇒ F=k\frac{q_{1}q_{2}}{r^{2}}$     -----(1)

Where k = 9 × 109 N-m2/C2, q1 and q2 = charges and r = distance between the charges

CALCULATION:

Given r = d

When the two charges are +6μC and +9μC, then the force F between them is given as,

$⇒ F=k\frac{q_{1}q_{2}}{r^{2}}$

$⇒ F=k\frac{6\times9}{d^{2}}$

$⇒ F=k\frac{54}{d^{2}}$     -----(1)

When a charge of -3μC is given to both the balls, the charges on the ball is given as,

⇒ q1 = +6 - 3

⇒ q1 = +3μC and,

⇒ q2 = +9 - 3

⇒ q2 = +6μC

• So the force between the charges become,

$⇒ F'=k\frac{q_{1}q_{2}}{r^{2}}$

$⇒ F'=k\frac{3\times6}{r^{2}}$

$⇒ F'=k\frac{18}{d^{2}}$     -----(2)

By equation 1 and equation 2,

$\Rightarrow \frac{F'}{F}=\frac{18}{54}$

$\Rightarrow F'=\frac{F}{3}$

• Hence, option 4 is correct.