RRB Technician Grade 1 Model Paper 1

Concept:The electric field due to a point charge is $E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}$; fields from multiple charges add vectorially.Explanation:Given $Q_1 = +1.8 \times 10^{-9}\ \mathrm{C}$ and $Q_2 = -1.8 \times 10^{-9}\ \mathrm{C}$.The separation is $6\ \mathrm{mm}$, so the midpoint is at $r = 3\ \mathrm{mm} = 3 \times 10^{-3}\ \mathrm{m}$ from each charge.Use $k = 9 \times 10^9\ \mathrm{N\,m^2\,C^{-2}}$.At the midpoint, the field due to $Q_1$ points away from the positive charge, and the field due to $Q_2$ points toward the negative charge.Since both charges lie on the x-axis, both fields point in the same direction at the midpoint.Thus, $E = E_1 + E_2 = \frac{k |Q_1|}{r^2} + \frac{k |Q_2|}{r^2}$.$E = \frac{9 \times 10^9 \times 1.8 \times 10^{-9}}{(3 \times 10^{-3})^2} + \frac{9 \times 10^9 \times 1.8 \times 10^{-9}}{(3 \times 10^{-3})^2}$.$E = 1.8 \times 10^6 + 1.8 \times 10^6 = 3.6 \times 10^6\ \mathrm{NC^{-1}}$.Therefore, the electric field at the midpoint is $3.6 \times 10^6\ \mathrm{NC^{-1}}$.Answer:$3.6 \times 10^6\ \mathrm{NC^{-1}}$ (Option D).