The volume integral \(\displaystyle \int_V e^{-\left(\frac{r}{R}\right)^2} \vec{\nabla} \cdot \left(\frac{\hat{r}}{r^2}\right) d^3r\), where \(V\) is the volume of a sphere of radius \(R\) centered at the origin, is equal to:
The volume integral \(\displaystyle \int_V e^{-\left(\frac{r}{R}\right)^2} \vec{\nabla} \cdot \left(\frac{\hat{r}}{r^2}\right) d^3r\), where \(V\) is the volume of a sphere of radius \(R\) centered at the origin, is equal to:
Show Hint
- \(4\pi\)
- \(0\)
- \(\dfrac{4}{3}\pi R^3\)
- \(1\)
The Correct Option is A
Solution and Explanation
Step 1: Evaluate divergence.
\[
\vec{\nabla} \cdot \left( \frac{\hat{r}}{r^2} \right) = 4\pi \delta^3(\vec{r})
\]
where \(\delta^3(\vec{r})\) is the Dirac delta function in three dimensions.
Step 2: Substitute into integral.
\[
\int_V e^{-(r/R)^2} \, 4\pi \delta^3(\vec{r}) \, d^3r = 4\pi e^{-(0/R)^2} = 4\pi
\]
Step 3: Conclusion.
Thus, the value of the volume integral is \(4\pi.\)
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