Let's solve the problem step by step: ## **Given:** The electrostatic potential is \[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3} \] where \(\mathbf{p}\) is a constant vector. --- ## **(1) Electric Field in Cartesian Coordinates** The electric field is given by: \[ \mathbf{E}(\mathbf{r}) = -\nabla V(\mathbf{r}) \] ### **Step 1: Compute the Gradient** Let \(\mathbf{p} \cdot \mathbf{r} = p_x x + p_y y + p_z z\) and \(r = |\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}\). So, \[ V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3} \] Take the gradient in Cartesian coordinates: \[ \nabla V = \frac{1}{4\pi\varepsilon_0} \nabla \left( \frac{\mathbf{p} \cdot \mathbf{r}}{r^3} \right) \] Let’s use the product rule: \[ \nabla [f(\mathbf{r})g(\mathbf{r})] = f \nabla g + g \nabla f \] Here, - \(f(\mathbf{r}) = \mathbf{p} \cdot \mathbf{r}\), so \(\nabla f = \mathbf{p}\) - \(g(\mathbf{r}) = 1/r^3\), so \(\nabla g = -3 r^{-4} \nabla r\), and \(\nabla r = \frac{\mathbf{r}}{r}\). Thus, \[ \nabla \left( \frac{1}{r^3} \right) = -3 r^{-4} \frac{\mathbf{r}}{r} = -3 \frac{\mathbf{r}}{r^5} \] So, \[ \nabla V = \frac{1}{4\pi\varepsilon_0} \left[ (\mathbf{p} \cdot \mathbf{r}) \left(-3 \frac{\mathbf{r}}{r^5}\right) + \frac{1}{r^3} \mathbf{p} \right] \] \[ = \frac{1}{4\pi\varepsilon_0} \left[ \frac{\mathbf{p}}{r^3} - 3 \frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} \right] \] ### **Step 2: Electric Field** \[ \mathbf{E}(\mathbf{r}) = -\nabla V(\mathbf{r}) = -\frac{1}{4\pi\varepsilon_0} \left[ \frac{\mathbf{p}}{r^3} - 3 \frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} \right] \] Or, \[ \boxed{ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \left[ 3 \frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} - \frac{\mathbf{p}}{r^3} \right] } \] This is the electric field of a point dipole in Cartesian form. --- ## **(2) Electric Field in Spherical Coordinates** Let’s choose the \(z\)-axis along the direction of \(\mathbf{p}\). - Then \(\mathbf{p} \cdot \mathbf{r} = p r \cos\theta\), where \(p = |\mathbf{p}|\). ### **Step 1: Express in Spherical Coordinates** - \(r = r\) - \(\cos\theta\) is the angle between \(\mathbf{p}\) and \(\mathbf{r}\). - \(\hat{r}\) and \(\hat{\theta}\) are the standard spherical unit vectors. The electric field in spherical coordinates: \[ \mathbf{E}(r, \theta) = \frac{1}{4\pi\varepsilon_0} \left[ 3 \frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} - \frac{\mathbf{p}}{r^3} \right] \] Let’s write both terms: - \(\mathbf{p} \cdot \mathbf{r} = p r \cos\theta\) - \(\mathbf{r} = r \hat{r}\) - \(\mathbf{p} = p \hat{z}\) Express \(\hat{z}\) in spherical coordinates: \[ \hat{z} = \cos\theta\,\hat{r} - \sin\theta\,\hat{\theta} \] So, \[ \mathbf{p} = p(\cos\theta\,\hat{r} - \sin\theta\,\hat{\theta}) \] The first term: \[ 3 \frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} = 3 \frac{p r \cos\theta \cdot r \hat{r}}{r^5} = 3p \frac{\cos\theta}{r^3} \hat{r} \] The second term: \[ \frac{\mathbf{p}}{r^3} = \frac{p}{r^3} (\cos\theta\,\hat{r} - \sin\theta\,\hat{\theta}) \] So, \[ \mathbf{E}(r, \theta) = \frac{1}{4\pi\varepsilon_0} \left[ 3p \frac{\cos\theta}{r^3} \hat{r} - \frac{p}{r^3} (\cos\theta\,\hat{r} - \sin\theta\,\hat{\theta}) \right] \] Expand: \[ = \frac{1}{4\pi\varepsilon_0} \left[ (3\cos\theta - \cos\theta) \frac{p}{r^3} \hat{r} + \sin\theta \frac{p}{r^3} \hat{\theta} \right] \] \[ = \frac{p}{4\pi\varepsilon_0 r^3} \left[ 2\cos\theta \hat{r} + \sin\theta \hat{\theta} \right] \] ### **Final Answer in Spherical Coordinates** \[ \boxed{ \mathbf{E}(r, \theta) = \frac{p}{4\pi\varepsilon_0 r^3} \left( 2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta} \right) } \] --- ### **Summary** **(1) Cartesian coordinates:** \[ \boxed{ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \left[ 3 \frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^5} - \frac{\mathbf{p}}{r^3} \right] } \] **(2) Spherical coordinates (with \(\mathbf{p}\) along \(z\)-axis):** \[ \boxed{ \mathbf{E}(r, \theta) = \frac{p}{4\pi\varepsilon_0 r^3} \left( 2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta} \right) } \]

The electric field of a dipole is derived from the potential by taking its gradient, resulting in a field that depends on the orientation and magnitude of the dipole moment. In spherical coordinates aligned along the dipole axis, the field exhibits a characteristic angular dependence, with radial and angular components proportional to \(\cos\theta\) and \(\sin\theta\), respectively.