Consider a dipole($q_{1}$ and $q_{2}$) with a separation distance of d, where $q_{1}$ is the positive charge and $q_{2}$ is the negative charge. A point b lies a distance x away from $q_{1}$ such that b, $q_{1}$ and $q_{2}$ form a straight line. Find the electric field at b and what happens if x becomes very large. Another point c lies a distance r from both $q_{1}$ and $q_{2}$, such that c, $q_{1}$ and $q_{2}$ forms a triangle with sides of length r, r and d. Find the electric field at c.

Using the principle of superposition of electric fields,

$$\begin{aligned} E_{b} &= \frac{q}{4 \pi \epsilon_{0} \left( x \, – \frac{d}{2} \right)^{2}} – \frac{q}{4 \pi \epsilon_{0} \left( x + \frac{d}{2} \right)^{2}} \\ &= \frac{q}{4 \pi \epsilon_{0} x^{2}} \left[ \left(1 \, – \frac{d}{2x} \right)^{-2} – \left(1 + \frac{d}{2x} \right)^{-2} \right] \end{aligned} $$

Taking x to be very large. Using binomial expansion where $\left( 1 – \frac{d}{2x} \right)^{-2} \approx 1 \, – (-2) \left( \frac{d}{2x} \right)$ and $\left(1 + \left(\frac{d}{2x} \right) \right)^{-2} \approx 1 + (-2) \left( \frac{d}{2x} \right)$,

$$\begin{aligned} E_{b} &= \frac{q}{4 \pi \epsilon x^{2}} \frac{2d}{x} \\ &= \frac{p}{2 \pi \epsilon_{0} x^{3}} \end{aligned}$$

Note that p is the electric dipole moment ($p = qd$) and $\vec{E}_{b}$ is in the $\vec{i}$ direction.

Done for $E_{b}$. For point c, notice the contribution to the electric field by the positive charge and negative charge. The vertical component of both contributions will cancel each other as the magnitude of the two charges and distance are the same. The electric field at point c will end up with only a horizontal component:

$$\vec{E}_{c} = \frac{2q \text{cos} \, \alpha}{4 \pi \epsilon_{0} r^{2}} \, \hat{i}$$

Next: Electric Field Of A Ring Of Charge

Please, edit the problem statement, “A point b lies a distance x away from q1 such that b, q1 and q2 form a straight line, ” it should be “A point b lies a distance x away from the middle point of the distance between q1 and q2, such that b, q1 and q2 form a straight line, “

Omar is right, otherwise the denominators of the first expression would look different.