(A) Given: Distance $d = 5 \, mm = 5 \times 10^{-3} \, m$,Potential difference $V = 50 \, V$,Mass $m = 10^{-15} \, kg$,Charge $q = 10^{-11} \, C$.Elec

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

(A) Given: Distance $d = 5 \, mm = 5 \times 10^{-3} \, m$,Potential difference $V = 50 \, V$,Mass $m = 10^{-15} \, kg$,Charge $q = 10^{-11} \, C$.Electric field $E = \frac{V}{d} = \frac{50}{5 \times 10^{-3}} = 10^4 \, V/m$.The force on the particle is $F = qE$.Using Newton's second law,$F = ma$,so $a = \frac{qE}{m}$.Substituting the values: $a = \frac{10^{-11} \times 10^4}{10^{-15}} = \frac{10^{-7}}{10^{-15}} = 10^8 \, m/s^2$.

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

Easy

Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

A
$10^8 \, m/s^2$
B
$5 \times 10^5 \, m/s^2$
C
$10^5 \, m/s^2$
D
$2 \times 10^3 \, m/s^2$

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

In space,the electric potential varies as $V = 20|\vec{r}|$ volt,where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ is the position vector. Then,the electric field in $N C^{-1}$ at the point $(4 \ m, 3 \ m, -5 \ m)$ is:

MediumAP EAMCET 2025

View Solution

Write the relation between electric field and electrostatic potential.

Medium

View Solution

Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)

EasyGUJCET 2010

View Solution

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

MediumAIPMT 2008

View Solution

In a certain region of space with volume $0.2 \ m^3$,the electric potential is found to be $5 \ V$ throughout. The magnitude of the electric field in this region is . . . . . . $N/C$.

EasyNEET 2020

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

Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

Similar Questions

In space,the electric potential varies as $V = 20|\vec{r}|$ volt,where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ is the position vector. Then,the electric field in $N C^{-1}$ at the point $(4 \ m, 3 \ m, -5 \ m)$ is:

Write the relation between electric field and electrostatic potential.

Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

In a certain region of space with volume $0.2 \ m^3$,the electric potential is found to be $5 \ V$ throughout. The magnitude of the electric field in this region is . . . . . . $N/C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module