Two point charges -3 - C and -8 - C repel each other with a force of 40 N- If a charge of -5 - C is added to each of them- then the force between them will becomeA- -10 NB- -10 NC- -20 ND- -20 N

The correct option is A 10 NIn second case, charges will be 2 μ C and +3 μ C Since F ∝ Q_1 Q_2 i.e. F/F'=Q_1 Q_2/Q'_1 Q'_2 ∴40/F'=3 × 8/ 2 × 3= 4 ⇒ F'=10 N ...

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Standard XII

Physics

Coulomb's Law - Mathematical Form

Two point cha...

Question

Two point charges $+ 3 μ C$ and $+ 8 μ C$ repel each other with a force of 40 N. If a charge of $- 5 μ C$ is added to each of them, then the force between them will become

-10 N

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+10 N

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+20 N

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-20 N

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Solution

The correct option is A -10 N
In second case, charges will be $+ 3 μ - 5 μ = - 2 μ C$ and $8 μ - 5 μ = + 3 μ C$ as we add $- 5 μ c$
Since $F \propto Q_{1} Q_{2} i . e . \frac{F}{F^{'}} = \frac{Q_{1} Q_{2}}{Q_{1}^{'} Q_{2}^{'}}$
$∴ \frac{40}{F^{'}} = \frac{3 \times 8}{- 2 \times 3} = - 4 \Rightarrow F^{'} = 10 N$ (Attractive)

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Two point charges +3μC and +8μC repel each other with a force of 40 N. If a charge of −5μC is added to each of them, then the force between them will become

The correct option is A -10 NIn second case, charges will be +3μ−5μ=−2μC and 8μ−5μ=+3μC as we add −5μc Since F∝Q1Q2i.e.FF′=Q1Q2Q′1Q′2 ∴40F′=3×8−2×3=−4⇒F′=10N (Attractive)

Two point charges $+ 3 μ C$ and $+ 8 μ C$ repel each other with a force of 40 N. If a charge of $- 5 μ C$ is added to each of them, then the force between them will become