Answer:
Here\[q=1C,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\] Number of lines of force \[=\] Electric flux \[=\frac{q}{{{\varepsilon }_{0}}}=\frac{1}{8.85\times {{10}^{-12}}}\] \[\mathbf{=1}\mathbf{.13\times 1}{{\mathbf{0}}^{\mathbf{11}}}\mathbf{.}\]
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