A point charge of \(+3\ \mathrm{nC}\) is placed at the point (3 cm, 0), and a point charge of \(+4\ \mathrm{nC}\) is placed at the point (0, 3 cm). The magnitude of the electric field at the origin is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(5.0\times10^{4}\ \mathrm{N\,C^{-1}}\)
The field at a point due to several charges is the vector sum of the fields of the separate charges. Each charge is 3 cm from the origin.
\( \begin{aligned} E_1 &= \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r^{2}} \\ &= \frac{9\times10^{9}\times3\times10^{-9}}{(0.03)^{2}} \\ &= 3\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
\( \begin{aligned} E_2 &= \frac{1}{4\pi\varepsilon_0}\frac{q_2}{r^{2}} \\ &= \frac{9\times10^{9}\times4\times10^{-9}}{(0.03)^{2}} \\ &= 4\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
\( E_1 \) points along the negative x-direction, away from the \( +3\ \mathrm{nC} \) charge. \( E_2 \) points along the negative y-direction. The two fields are at right angles.
\( \begin{aligned} E &= \sqrt{E_1^{2}+E_2^{2}} \\ &= \sqrt{(3\times10^{4})^{2}+(4\times10^{4})^{2}} \\ &= 5.0\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Electric field due to a system of charges – “By the superposition principle, the electric field E at r due to the system of charges is … E(r) = E1 (r) + E2 (r) + … + En(r)”