SolveIt is under development
SolveItJEE Main
Physics · 2019

JEE Main · 9 January 2019, Shift 2 · Q16

Two point charges q_1(√ 10 μ C ) and q_2(-25 μ C ) are placed on the x -axis at x=1 m and x=4 m respectively. The electric field (in V / m ) at a…

Two point charges $\displaystyle \mathrm{q}_1(\sqrt{10} \mu \mathrm{C})$ and $\displaystyle \mathrm{q}_2(-25 \mu \mathrm{C})$ are placed on the $\displaystyle x$-axis at $\displaystyle x=1 \mathrm{~m}$ and $\displaystyle x=4 \mathrm{~m}$ respectively. The electric field (in $\displaystyle \mathrm{V} / \mathrm{m}$ ) at a point $\displaystyle y=3 \mathrm{~m}$ on $\displaystyle y$-axis is, $$\left[\text { take } \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right] $$
ShareWhatsAppTelegram

More from Electrostatics

JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.