The force between two bodies having charges $5 C$ and $10 C$ separated by distance of $0.5 m$ in air is :

18 \times 10^{11} N

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9 \times 10^{11} N

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9 \times 10^{9} N

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18 \times 10^{9} N

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Solution

The correct option is A $18 \times 10^{11} N$
By Coulomb's law, the force between two charges is $F = k \frac{q_{1} q_{2}}{r^{2}} = (9 \times 10^{9}) \times \frac{5 \times 10}{(0.5)^{2}} = 18 \times 10^{11} N$

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Electric charges of $1 μ C, - 1 μ C$ and $2 μ C$ are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is $(\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} N - m^{2} / c^{2})$ :

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The force between two bodies having charges 5C and 10C separated by distance of 0.5m in air is :

The correct option is A 18×1011 NBy Coulomb's law, the force between two charges is F=kq1q2r2=(9×109)×5×10(0.5)2=18×1011N

The force between two bodies having charges $5 C$ and $10 C$ separated by distance of $0.5 m$ in air is :

The correct option is A $18 \times 10^{11} N$
By Coulomb's law, the force between two charges is $F = k \frac{q_{1} q_{2}}{r^{2}} = (9 \times 10^{9}) \times \frac{5 \times 10}{(0.5)^{2}} = 18 \times 10^{11} N$