Question

Let \( \mathbf{a}_R \) be the unit radial vector in the spherical coordinate system. For which of the following value(s) of \( n \), the divergence of the radial vector field \( \mathbf{f}(R) = \mathbf{a}_R \frac{1}{R^n} \) is independent of \( R \)?

• A. \( -2 \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

For a radial field \( \mathbf{f}(R) = \mathbf{a}_R \frac{1}{R^n} \), use: \[ \nabla \cdot \mathbf{f} = \frac{1}{R^2} \frac{d}{dR} \left(R^2 \cdot \frac{1}{R^n} \right) = (2 - n) R^{-1 - n} \] To find when divergence is {independent of \( R \)}, solve for when the exponent is zero: \( -1 - n = 0 \Rightarrow n = -1 \). Also, divergence is zero for \( n = 2 \), which is constant too.

Answer 1

The Correct Option is B, D

In spherical coordinates, for a vector field \( \mathbf{f}(R) = \mathbf{a}_R F(R) \), the divergence is: \[ \nabla \cdot \mathbf{f} = \frac{1}{R^2} \frac{d}{dR} \left( R^2 F(R) \right) \] Given \( F(R) = \frac{1}{R^n} \), we get: \[ \nabla \cdot \mathbf{f} = \frac{1}{R^2} \frac{d}{dR} \left( R^2 \cdot \frac{1}{R^n} \right) = \frac{1}{R^2} \cdot \frac{d}{dR} \left( R^{2-n} \right) = \frac{1}{R^2} \cdot (2 - n) R^{1 - n} = (2 - n) R^{-1 - n} \] For the divergence to be independent of \( R \), the exponent of \( R \) must be zero: \[ -1 - n = 0 \Rightarrow n = -1 \] Now test if there are other such values. Let’s try the expression: \[ \nabla \cdot \mathbf{f} = (2 - n) R^{-1 - n} \] This will be independent of \( R \) if the exponent is zero: \[ -1 - n = 0 \Rightarrow n = -1 \] So, only \( n = -1 \) strictly satisfies divergence being constant. However, if we want zero divergence, then: \[ \nabla \cdot \mathbf{f} = 0 \Rightarrow (2 - n) R^{-1 - n} = 0 \Rightarrow 2 - n = 0 \Rightarrow n = 2 \] So, for: - \( n = -1 \): divergence is constant (independent of \( R \)) - \( n = 2 \): divergence is zero, which is also independent of \( R \) 
Hence, both \( n = -1 \) and \( n = 2 \) are correct. \[ \boxed{{Correct options: (B), (D)}} \]