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(A) The electric field $E$ produced by a point charge $Q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{

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$8 	imes 10^{-11} \, C$

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(A) The electric field $E$ produced by a point charge $Q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r^2}$.Given values are $E = 2 \, N/C$,$r = 60 \, cm = 0.6 \, m$,and $k = \frac{1}{4\pi \varepsilon_0} = 9 	imes 10^9 \, N \cdot m^2/C^2$.Substituting these values into the formula:$2 = (9 	imes 10^9) \cdot \frac{Q}{(0.6)^2}$$2 = (9 	imes 10^9) \cdot \frac{Q}{0.36}$$Q = \frac{2 	imes 0.36}{9 	imes 10^9}$$Q = \frac{0.72}{9 	imes 10^9} = 0.08 	imes 10^{-9} \, C = 8 	imes 10^{-11} \, C$.Thus,the magnitude of the point charge is $8 	imes 10^{-11} \, C$.

What is the magnitude of a point charge which produces an electric field of $2 \, N/C$ at a distance of $60 \, cm$? (Given: $1/(4\pi \varepsilon_0) = 9 \times 10^9 \, N \cdot m^2/C^2$)

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