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Question

(A) The electric field $E$ due to a point charge $q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}$

Vedclass · Accepted Answer

$2 	imes 10^{-11}\,C$

Vedclass · Answer

(A) The electric field $E$ due to 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: $E = 2\,N/C$,$r = 30\,cm = 0.3\,m$,and $\frac{1}{4\pi \varepsilon_0} = 9 	imes 10^9\,N\cdot m^2/C^2$.Rearranging the formula to solve for $q$: $q = E \cdot r^2 \cdot (4\pi \varepsilon_0) = \frac{E \cdot r^2}{1/(4\pi \varepsilon_0)}$.Substituting the values: $q = \frac{2 	imes (0.3)^2}{9 	imes 10^9} = \frac{2 	imes 0.09}{9 	imes 10^9} = \frac{0.18}{9 	imes 10^9} = 0.02 	imes 10^{-9}\,C = 2 	imes 10^{-11}\,C$.

What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,N/C$ $[1/4\pi \varepsilon_0 = 9 \times 10^9\,N\cdot m^2/C^2]$?

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