Two point charges q_A = 3 μC and q_B = –3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10^–9 C is placed at this point, what is the force experienced by the test charge?

Given,
                          $$\begin{gathered} \mathrm{Magnitude} \mathrm{of} \mathrm{charge} \mathrm{A}-{\mathrm{q}}_{\mathrm{A}} = 3 \mathrm{\mu C} = 3 \times {10}^{-6}\mathrm{C} \\ \mathrm{Magnitude} \mathrm{of} \mathrm{charge} \mathrm{B}-{\mathrm{q}}_{\mathrm{B}} = -3 \mathrm{\mu C} = -3 \times {10}^{-6}\mathrm{C} \end{gathered}$$

Distance between two charges-d= 20 cm

Distance of charge A and B from point O - r= 20/2 = 10cm =0.1 m


(a) 
         

Let us assume that a unit positive test charge is placed at 0. q_A will repel this test charge while q_B will attract. Hence, $\overrightarrow{{\mathrm{E}}_1} \mathrm{and} \overrightarrow{{\mathrm{E}}_2}$ both are directed towards $\overrightarrow{\mathrm{OB}}.$
$\therefore$                           $\mathrm{E} = \overrightarrow{{\mathrm{E}}_1} + \overrightarrow{{\mathrm{E}}_2}$

                               $\boxed{\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}}$
                               
                                     $$\begin{gathered} = \frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{{\mathrm{q}}_{\mathrm{A}}}{{\mathrm{r}}^2}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_{\mathrm{B}}}{{\mathrm{r}}^2} =\frac{1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left[{\mathrm{q}}_{\mathrm{A}}+{\mathrm{q}}_{\mathrm{B}}\right] \\ = \frac{9\times {10}^9}{(0.1{)}^2}\left[3\times {10}^{-6}+3\times {10}^{-6}\right] \\ = 5.4 \times {10}^6 {\mathrm{NC}}^{-1} \mathrm{along} \mathrm{OB}. \end{gathered}$$

(b) As a negative test charge of q₀ = – 1.5 x 10^–6 C is placed at 0. q_A will attract it while q_B will repel. 


Therefore, the net force

$$\begin{gathered} \mathrm{F} = {\mathrm{F}}_1+{\mathrm{F}}_2 \\ \boxed{\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}} \end{gathered}$$
$$\begin{gathered} \mathrm{F}=\left|{\mathrm{F}}_1\right|+\left|{\mathrm{F}}_2\right| \\ = \frac{{\mathrm{Kq}}_{\mathrm{A}}{\mathrm{q}}_0}{{\mathrm{r}}^2}+\frac{{\mathrm{Kq}}_{\mathrm{B}}{\mathrm{q}}_0}{{\mathrm{r}}^2} \\ = \frac{{\mathrm{K}}_{{\mathrm{q}}_0}}{{\mathrm{r}}^2}\left[{\mathrm{q}}_{\mathrm{A}}+{\mathrm{q}}_0\right] \\ = \frac{9\times {10}^9\times 1.5\times {10}^{-19}[3\times {10}^{-6}+3\times {10}^{-6}]}{(0.1{)}^2} \end{gathered}$$
   $$\begin{gathered} = \frac{9\times 1.50\times 6\times {10}^{-6+9-9}}{0.1\times 0.1} \\ = \frac{9\times {10}^9\times 3\times {10}^{-6}\times 1.5\times {10}^{-9}}{(0.1{)}^2}+\frac{9\times {10}^9\times 3\times {10}^{-6}\times 1.5\times {10}^{-9}}{(0.1{)}^2} \\ = \frac{9\times {10}^9\times 3\times {10}^{-6}\times 1.5\times {10}^{-9}\times 2}{(0.1{)}^2} \\ = 8.1 \times {10}^{-3}\mathrm{N}. \mathrm{along} \mathrm{OA} \end{gathered}$$