A point charge q = l $$\mu C$$ is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge Q = 24 $$\mu C$$, distributed uniformly along its length, as shown in figure. Force between q and wire is __ N.
(Use: $$\frac{1}{4\pi\epsilon_{0}}=9 \times10^{9} N.m^{2}/C^{2}$$)

Wire length

L=10 cm=0.1mCharge on wire

$$Q=24\mu C=24\times10^{-6}C$$

Point charge

$$q=1\mu C=10^{-6}C$$

Distance of charge from nearer end:

a=2 cm=0.02m

Far end distance:

$$a+L=0.12m$$

Linear charge density

$$\lambda=\frac{Q}{L}=\frac{24\times10^{-6}}{0.1}=2.4\times10^{-4}C/m$$

For small element dx,

$$dF=\frac{1}{4\pi\epsilon_0}\frac{q\lambda dx}{x^2}$$

Integrate from x=0.02 to 0.12:

$$F=9\times10^9(10^{-6})(2.4\times10^{-4})\int_{0.02}^{0.12}\frac{dx}{x^2}=9\times10^9(2.4\times10^{-10})\left[-\frac{1}{x}\right]_{0.02}^{0.12}$$

$$=2.16\left(\frac{1}{0.02}-\frac{1}{0.12}\right)$$

$$=2.16(50-8.33)$$

$$=2.16(41.67)$$

=90