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 \_\_\_\_ $\displaystyle \hat{i} \mathrm{~V} / \mathrm{m}$.
Official answer
5500
More from Electrostatics
- The vehicles carrying inflammable fluids usually have metallic chains touching the ground:2024
- A point charge of +12 μ C is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the…2021
- An electron revolves around an infinite cylindrical wire having uniform linear charge density 2 × 10^-8 C m^-1 in circular path under the influence…2023
- Six point charges are kept 60^° apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the…2026
- A particle of charge ' -q ' and mass ' m ' moves in a circle of radius ' r ' around an infinitely long line charge of linear charge density ' +λ '.…2024
- Given below are two statements: one is labelled as Assertion and the other is labelled as Reason (R). Assertion: Work done by electric field on…2024
- Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1,…2020
- Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ and…2025
JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.