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Physics Question on Energy in simple harmonic motion

Two point charges 3×10−6C3\times {{10}^{-6}}C and 8×10−6C8\times {{10}^{-6}}C repel each other by a force of 6×10−3N6\times {{10}^{-3}}N . If each of them is given an additional charge −6×10−6C-6\times {{10}^{-6}}C , the force between them will be:

A

2.4×10−3N2.4\times {{10}^{-3}}N (repulsive)

B

2.4×10−3N2.4\times {{10}^{-3}}N (attractive)

C

1.5×10−3N1.5\times {{10}^{-3}}N (repulsive)

D

1.5×10−3N1.5\times {{10}^{-3}}N (attractive)

Answer

1.5×10−3N1.5\times {{10}^{-3}}N (attractive)

Explanation

Solution

Like charges repel each other while unlike charges attract each other.
From Coulombs law, the force of attraction/repulsion between two point charges q1q_{1} and q2q_{2} placed a distance rr apart is given by
F=14πε0q1q2r2NF=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} N
When similar charges are taken
q1=3×10−6C,q2=8×10−6Cq_{1}=3 \times 10^{-6} C , q_{2}=8 \times 10^{-6} C
F=14πε0(3×10−6)×(8×10−6)r2?F =\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(3 \times 10^{-6}\right) \times\left(8 \times 10^{-6}\right)}{r^{2}} ? (i) (repulsive)
When additional charge −6×10−6C-6 \times 10^{-6} C is given to each charge, then
F=14πε0(3−6)×10−6×(8−6)×10−6r2F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(3-6) \times 10^{-6} \times(8-6) \times 10^{-6}}{r^{2}} (attractive)
∴F=14πε0(−3)×10−6×2×10−6r2N?\therefore F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(-3) \times 10^{-6} \times 2 \times 10^{-6}}{r^{2}} N ? (ii)
Dividing E (ii) by E (i), we get
FF=−624⇒F=−F4\frac{F}{F}=-\frac{6}{24} \Rightarrow F=-\frac{F}{4}
=−6×10−34=−1.5×10−3N=-\frac{6 \times 10^{-3}}{4}=-1.5 \times 10^{-3} N
Negative sign indicates, force is attractive.