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Question: Two point charges 3 × 10<sup>–6</sup> C and 8 × 10<sup>–6</sup> C repel each other by a force of 6 ×...

Two point charges 3 × 10–6 C and 8 × 10–6 C repel each other by a force of 6 × 10–3 N. If each of them is given an additional charge – 6 × 106 C, the force between them will be

A

2.4 × 10–3 N (attractive)

B

2.4 × 10–9 N (attractive)

C

1.5 × 10–3 N (repulsive)

D

1.5 × 10–3 N (attractive)

Answer

1.5 × 10–3 N (attractive)

Explanation

Solution

F∝Q1Q2F \propto Q_{1}Q_{2} ⇒ F1F2=Q1Q2Q1′Q2′\frac{F_{1}}{F_{2}} = \frac{Q_{1}Q_{2}}{Q_{1}'Q_{2}'}

=3×10−6×8×10−6(3×10−6−6×10−6)(8×10−6−6×10−6)=3×8−3×2=−41= \frac{3 \times 10^{- 6} \times 8 \times 10^{- 6}}{(3 \times 10^{- 6} - 6 \times 10^{- 6})(8 \times 10^{- 6} - 6 \times 10^{- 6})} = \frac{3 \times 8}{- 3 \times 2} = - \frac{4}{1}

⇒ F2=−F14=−6×10−34=−1.5×10−3NF_{2} = - \frac{F_{1}}{4} = - \frac{6 \times 10^{- 3}}{4} = - 1.5 \times 10^{- 3}N (Attractive)