SolveIt is under development
SolveItJEE Main
Question 166 / 212 · latest first
Physics · 2021

JEE Main · 3 August 2021, Shift 2 · Q29

The electric potential at point (x, 0,0) is V=(1000/x+1500/x^2+500/x^3) volt. The electric field strength at a point x=1 m is \ \ \ \ i V / m.

The electric potential at point $\displaystyle (x, 0,0)$ is $\displaystyle V=\left(\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3}\right)$ volt. The electric field strength at a point $\displaystyle x=1 \mathrm{~m}$ is \_\_\_\_ $\displaystyle \hat{i} \mathrm{~V} / \mathrm{m}$.
ShareWhatsAppTelegram

More from Electrostatics

JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.