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प्रश्न
The electric field due to a point charge Q at a distance r is given by:
विकल्प
\[\vec{E}=\frac{1}{2\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}\]
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}\]
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q^2}{r^2}\hat{r}\]
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}\hat{r}\]
MCQ
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उत्तर
The electric field due to a point charge Q is \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}\]. This follows from Coulomb's law divided by the test charge, giving force per unit charge with an inverse-square dependence on distance r.
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