Two point charges q₁ = +0.2 C and q₂ = +0.4 C are placed 0.1 m apart. Calculate the electric field at (a) the mid-point between the charges and (b) at a point on the line joining q₁ and q₂ such that it is 0.05 m from q₂ and 0.15 m away from q₁.

Given, q₁= +0.2 C
         q₂= +0.4 C
distance between the charges-d=0.1 m 

a)Electric field at the mid point between these two charges:



Electric field due to q₁-E₁=
          $=9\times {10}^9\times \frac{0.2}{0.{05}^2}=720\times {10}^{9 } \mathrm{N}$/C
Electric field due to q2-E₂=$\frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }} \frac{0.4}{0.{05}^2}$
          $=9\times {10}^9\times \frac{0.4}{0.{05}^2}=1440\times {10}^{9 } \mathrm{N}$/C

Resultant Electric field at mid-point-E=${\stackrel{\rightharpoonup }{\mathrm{E}}}_1+{\stackrel{\rightharpoonup }{E}}_2$
Since the net electric field is acting in opposite direction we have E= 1440$\times {10}^9-720\times {10}^{9 } \mathrm{N}$
                           = $720\times {10}^{9 } \mathrm{N}$/C

b) 

Let point P be on the line joining the charges such that it is 0.05m away from q₂ and 0.15 m away from q_1.

Electric field due to q_{1= $9\times {10}^9\times \frac{0.2}{0.{15}^2}=80\times {10}^{9 }N/C$}
Electric field due to q₂=$9\times {10}^9\times \frac{0.4}{0.{05}^2}=1440\times {10}^{9 }N$/C
Since, Electric field is acting in the same direction

Resultant electric field intensity-E=$$\begin{gathered} 80\times {10}^9 + 1440\times {10}^9 \\ =1520\times {10}^9 \\ =15.2\times {10}^{11 }\mathrm{N}/\mathrm{C} \end{gathered}$$