(A) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.Given $V = 5x^2 + 10x - 9$.Dif

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.Given $V = 5x^2 + 10x - 9$.Differentiating $V$ with respect to $x$,we get $\frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) = 10x + 10$.Substituting this into the electric field formula: $E = -(10x + 10)$.At $x = 1 \, m$,the electric field is $E = -(10(1) + 10) = -(10 + 10) = -20 \, V m^{-1}$.

Class 12 Physics Electric Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

Easy

The 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

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

Subtopic

Relation between Electric Field and Potential and Potential Gradient

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + y^2\hat j$. If the potential at $(0, 0, 0)$ is $2 \, V$,find the potential at $(1, 1, 1)$. (in $/3$)

Medium

View Solution

For a given electric field $\vec{E} = 2x\hat{i} + 3y\hat{j}$,find the potential at $(X, Y)$ if the potential at the origin is $5\, V$.

Medium

View Solution

The potential $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

EasyTS EAMCET 2018

View Solution

The electric field at a distance $x$ from the origin is given by $E = 100/x^2$. The potential difference between the points at $x = 10 \, m$ and $x = 20 \, m$ is ...... $V$.

Difficult

View Solution

Assume that an electric field $E = 30x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x = 2 \ m$,then the potential difference $(V_A - V_0)$ is: (in $V$)

MediumMHT CET 2023

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The 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 ...... $V m^{-1}$.

Similar Questions

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + y^2\hat j$. If the potential at $(0, 0, 0)$ is $2 \, V$,find the potential at $(1, 1, 1)$. (in $/3$)

For a given electric field $\vec{E} = 2x\hat{i} + 3y\hat{j}$,find the potential at $(X, Y)$ if the potential at the origin is $5\, V$.

The potential $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

The electric field at a distance $x$ from the origin is given by $E = 100/x^2$. The potential difference between the points at $x = 10 \, m$ and $x = 20 \, m$ is ...... $V$.

Assume that an electric field $E = 30x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x = 2 \ m$,then the potential difference $(V_A - V_0)$ is: (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module