English: So, we can compare this with those answers, which is given by r over three epsilon naught cos theta; for r is smaller than capital R. and, equal to R cube over three epsilon naught cos theta over r square; for r is greater than equal to capital R. So, we can see that; we can write this as cos theta prime sin theta prime d theta prime d phi prime over vector r minus vector R prime one over four pi epsilon naught R square equal to r over 3 epsilon naught cos theta; for r is smaller than equal to capital R. And, equal to R cube over three epsilon naught cos theta over r square; for r is greater than equal to capital R. $\frac{R^{2}}{4\pi\epsilon_{0}}\ \frac{\cos\theta^{'}}{\left| \overrightarrow{r} - \overrightarrow{R^{'}} \right|}\sin\theta^{'}d\theta^{'}d\phi^{'} = \ \frac{r}{3\epsilon_{0}}\ cos\theta\ \ \ \ for\ \ r \leq R\ \ \ \ \ \ \ \ \ \ \ = \frac{R^{3}}{3\epsilon_{0}}\ \frac{cos\theta}{r^{2}}\ \ \ \ \ for\ \ r \geq R\frac{\cos\theta^{'}}{\left| \overrightarrow{r} - \overrightarrow{R^{'}} \right|}\sin\theta^{'}d\theta^{'}d\phi^{'} = \ \frac{4\pi r}{3R^{2}}cos\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ r \leq R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{4\pi R}{3r^{2}}cos\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ r \geq R$ $\frac{\cos\theta^{'}\sin\theta^{'}d\theta^{'}d\phi^{'}}{}$ So, you can write this as sin theta prime cos theta prime d theta prime d phi prime over vector r minus vector R prime equal to; so you can see, this can be written as r over capital R square into 4 pi cos theta; for r is smaller than capital R. And, equal to 4 pi R over 3 cos theta over r square; for r is greater than equal to capital R. So here in a spherical coordinate, vector r minus vector capital R can be written equal to, sorry mod of vector, mod of vector r minus vector capital R small r cos theta prime cos theta plus sin theta prime sin theta cos phi minus phi prime.
Telugu: |r⃗−R⃗| = r² + R² − 2rR(cosθ^(′)cosθ+sinθ^(′)sinθcos(ϕ−ϕ′)) can be written equal to r square plus capital R square minus 2 capital R phi prime over r square plus capital R square minus 2 r capital R cos కాబట్టి, మీరు దీనిని సైన్ తీటా ప్రైమ్ కాస్ తీటా ప్రైమ్ డి తీటా ప్రైమ్ డి ఫై ప్రైమ్ ఓవర్ వెక్టర్ ఆర్ మైనస్ వెక్టర్ ఆర్ ప్రైమ్‌కి సమానంగా వ్రాయవచ్చు; కాబట్టి మీరు చూడగలరు, దీనిని ఆర్ పై క్యాపిటల్ R స్క్వేర్‌గా 4 పై కాస్ తీటాగా వ్రాయవచ్చు; r అనేది క్యాపిటల్ R.