An electric field converges at the origin who | vedclass

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(D) Since the electric field converges at the origin,the electric field vector $\vec{E}$ is directed towards the origin. Thus,$\vec{E} = -100\,r\hat{r

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All of the above.

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(D) Since the electric field converges at the origin,the electric field vector $\vec{E}$ is directed towards the origin. Thus,$\vec{E} = -100\,r\hat{r}$.By Gauss's Law,$\oint \vec{E} \cdot d\vec{s} = \frac{q_{enclosed}}{\epsilon_0}$.For a spherical surface of radius $r$ centered at the origin,the flux is $\Phi = E \cdot (4\pi r^2) = (-100r)(4\pi r^2) = -400\pi r^3$.Since the flux is negative,the enclosed charge $q$ must be negative,which confirms that the charge contained in any spherical volume centered at the origin is negative.For $r = 3\,cm = 0.03\,m$,the charge $q = \epsilon_0 \Phi = (8.854 	imes 10^{-12}) 	imes (-400\pi 	imes (0.03)^3) \approx -3 	imes 10^{-13}\,C$.Thus,all the given statements are correct.

An electric field converges at the origin whose magnitude is given by the expression $E = 100\,r\,N/C$,where $r$ is the distance measured from the origin.

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