An electric dipole (discussed in detail in Section 4.7) consists of two point charges of equal and opposite magnitude \(\pm Q\) spaced by distance \(d\). With the charges along the \(z\) axis at positions \(z=\pm d / 2\) (with the positive charge at the positive \(z\) location), the electric field in spherical coordinates is given by \(\mathbf{E}(r, \theta)=\left[Q d /\left(4 \pi \epsilon_{0} r^{3}\right)\right]\left[2 \cos \theta \mathbf{a}_{r}+\sin \theta \mathbf{a}_{\theta}\right]\), where \(r>>d\). Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude \(q(a)\) at \((0,0, z) ;(b)\) at \((0, y, 0)\).

Given the electric field in spherical coordinates is: \(\mathbf{E}(r, \theta)=\left[\frac{Q d}{4 \pi \epsilon_{0} r^{3}}\right]\left[2 \cos \theta \mathbf{a}_{r} + \sin \theta \mathbf{a}_{\theta}\right]\)

We need to find the electric field at position (0, 0, z) in rectangular coordinates. Since we are given \(\mathbf{E}(r,\theta)\) in spherical coordinates, we must convert this expression to rectangular coordinates using the relations: \(r = \sqrt{x^2+y^2+z^2}, \ \ \ \theta = \cos^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\) We will also need the transformation from unit vectors in spherical coordinates to rectangular coordinates: $\mathbf{a}_r = \sin\theta \cos\phi \ \mathbf{a}_x + \sin\theta \sin\phi \ \mathbf{a}_y + \cos\theta \mathbf{a}_z \\ \mathbf{a}_\theta = \cos\theta \cos\phi \ \mathbf{a}_x + \cos\theta \sin\phi \ \mathbf{a}_y - \sin\theta \mathbf{a}_z \\ (\phi = 0\ , \textrm{since we are looking at the x-z plane})$ Now substitute the position (0, 0, z) in the given 𝑬(𝑟,𝜃) and transform the spherical coordinates to rectangular coordinates: \(r = \sqrt{z^2} = |z|, \ \ \theta = \cos^{-1}\left(\frac{z}{|z|}\right)\) Plugging these into the electric field expression and converting the unit vectors, we get: \(\mathbf{E}(0,0,z) = \left[\frac{Q d}{4 \pi \epsilon_{0} |z|^3}\right]\left[2 \frac{z}{|z|}\mathbf{a}_{z} + \left(\frac{z}{|z|}\mathbf{a}_x - \frac{z^2}{|z|^2}\mathbf{a}_z \right)\right]\)

We know that the force on a point charge is given by: \(\mathbf{F} = q \mathbf{E}\) So the force acting on charge \(q\) placed at (0, 0, z) is: \(\mathbf{F}(0, 0, z) = q\mathbf{E}(0, 0, z) = q \left[\frac{Q d}{4 \pi \epsilon_{0} |z|^3}\right]\left[2 \frac{z}{|z|}\mathbf{a}_{z} + \left(\frac{z}{|z|}\mathbf{a}_x - \frac{z^2}{|z|^2}\mathbf{a}_z \right)\right]\). This is the expression for the force vector on the point charge at (0, 0, z) in rectangular coordinates.

Now, we need to find the electric field at position (0, y, 0) in rectangular coordinates. We will use the same relations for the conversion, but in this case: \(r = \sqrt{y^2} = |y|, \ \ \theta = \cos^{-1}\left(\frac{-d/2}{\sqrt{y^2+(d/2)^2}}\right)\) Plugging these into the electric field expression and converting the unit vectors, we get: \(\mathbf{E}(0, y, 0) = \left[\frac{Q d }{4 \pi \epsilon_{0} |y|^3}\right]\left[2 \frac{(-d/2)\mathbf{a}_z}{\sqrt{y^2 + (d/2)^2}} + \frac{y\mathbf{a}_y}{\sqrt{y^2 + (d/2)^2}}\right]\)

Using the given expression for force on a point charge, we can find the force acting on charge \(q\) placed at (0, y, 0): \(\mathbf{F}(0, y, 0) = q\mathbf{E}(0, y, 0) = q \left[\frac{Q d}{4 \pi \epsilon_{0} |y|^3}\right]\left[2 \frac{(-d/2)\mathbf{a}_z}{\sqrt{y^2 + (d/2)^2}} + \frac{y\mathbf{a}_y}{\sqrt{y^2 + (d/2)^2}}\right]\) This is the expression for the force vector on the point charge at (0, y, 0) in rectangular coordinates.