Class 12PhysicsElectrostatic Potential and CapacitanceRelation between Electric Field and Potential and Potential Gradient
EasyThe electric potential at any point as a function of distance $(x)$ in meters is given by $V = 5x^2 + 10x - 9 \, (V)$. The value of the electric field at $x = 1 \, m$ is ...... $V m^{-1}$.
- A-$20$
- B$6$
- C$11$
- D-$23$
Explore More
Similar Questions
The variation of potential $V$ with distance $x$ from a fixed point is as shown in the figure. The electric field at $x = 13\,m$ is......$V/m$.
Medium
View SolutionIf on the $x$-axis electric potential decreases uniformly from $60 \,V$ to $20 \,V$ between $x=-2 \,m$ to $x=+2 \,m$,then the magnitude of electric field at the origin
Medium
View SolutionThe potential $V$ varies with $x$ and $y$ as $V = \frac{1}{2}(y^2 - 4x) \text{ volts}$. The electric field at $(1 \text{ m}, 1 \text{ m})$ is:
Difficult
View SolutionThe figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?
Medium
View SolutionIn which region is the magnitude of the $x$-component of the electric field maximum,if the potential $(V)$ versus distance $(X)$ graph is as shown?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo