Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

A uniform electric field  of $12$ $V/m$ is along the positive $x$ direction. Determine the potential difference in volts, between $x=0m$ and $x=3m$.

  1. $-27$ $V$

  2. $-36$&nbsp;<span>$V$</span>

  3. $27$&nbsp;<span>$V$</span>

  4. $36$&nbsp;<span>$V$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
As we know, relation between uniform electric field strength $\&$ potential difference between two point a distance d
$V=-Ed \; \Rightarrow \; V=-(12)(3)=-36 \; Volts$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

In the direction of electric field, the electric potential:

  1. decreases

  2. increases

  3. remains uncharged

  4. becomes zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of electric field.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Variation of potential V with distance r in electric field of E$=0$ is?

  1. $V\propto \displaystyle\frac{1}{r}$

  2. $V\propto r$

  3. $V\propto \displaystyle\frac{1}{r^2}$

  4. $V=$ constant
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Potential difference between two points is given by    $\Delta V = -E.r$
Given :  $E = 0$
$\implies  \ \Delta V = 0$
$\implies \ V =$ constant

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electric potential decreases uniformly from V to -V along X-axis in a coordinate system as we moves from a (-$x _0$, 0) to ($x _0$, 0), then the electric field at the origin.

  1. must be equal to $\dfrac{V}{x _0}$;

  2. may be equal to $\dfrac{V}{x _0}$;

  3. must be greater than $\dfrac{V}{x _0}$;

  4. may be less than $\dfrac{V}{x _0}$;
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

A copper ball of radius 1 cm work function 4.47 eV is irradiated with ultraviolet radiation of wavelength $2500\mathring { A } $. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be finite value of the potential on the ball. The charge acquired by the ball is :

  1. $5.5\times { 10 }^{ -13 }C$

  2. $7.5\times { 10 }^{ -13 }C$

  3. $4.5\times { 10 }^{ -12 }C$

  4. $2.5\times { 10 }^{ -11 }C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From photo electric effect equation :

]
$h\nu=h\nu _{0} + K.E _{max}$


so Maximum kinetic energy will be


$K.E _{max}= \dfrac{hc}{\lambda} - h\nu _{0}$
 
putting the given values in the above equation

$K.E _{max} = e\times V$ 

so V will be 

$V= \dfrac{k\times Q}{r}$
 
:: $ q = 5.5\times 10^{-13} C $

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Two infinite, parallel, non-conducting sheets carry equal positive charge density $\sigma$. One is placed in the yz plane at $x=0$ and the other at distance $x=a$. Take potential $V=0$ at $x=0$. Then,

  1. for $0\leq x \leq a$, potential $V _x=0$

  2. for $x\geq a$, potential $V _x=-\frac {\sigma}{\epsilon _0}(x-a)$

  3. for $x\geq a$, potential $V _x=\frac {\sigma}{\epsilon _0}(x-a)$

  4. for $x\leq 0$ potential $V _x=\frac {\sigma}{\epsilon _0}x$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Now , Since both are infinite plates and carry same charge density therefore, electric field between them will be equal to zero.
Now, potential will be constant between them and at $x=0; V=0$ and V = constant between the plates.
Therefore, V=0 between the plates means $0\le x\le a$
Now electric field beyond $x=a$ is $2\times \sigma/2\epsilon _o=\sigma/\epsilon _o$
We know that,
$V=-\int _{ a }^{ x }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=-E(x-a)$
$V=-\sigma(x-a)/\epsilon _o$

and for $x<0$
$V=-\int _{ x }^{ 0 }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=Ex$
$V=\sigma x/\epsilon _o$
option (A)(B)(D) are correct.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Electric potential $'v'$ in space as a function of co-ordinates is given by, $v=\cfrac{1}{x}+\cfrac{1}{y}+\cfrac{1}{z}$. Then the electric field intensity at $(1,1,1)$ is given by :

  1. $-(\hat { i } +\hat { j } +\check { k } )$

  2. $\hat { i } +\hat { j } +\check { k } $

  3. zero

  4. $\cfrac{1}{\sqrt 3}(\hat { i } +\hat { j } +\check { k } )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The electric field , $\vec{E}=-\vec{\nabla}V=-\left[\dfrac{\partial V}{\partial x}\hat i+\dfrac{\partial V}{\partial y}\hat j+\dfrac{\partial V}{\partial z}\hat k\right]=\dfrac{1}{x^2}\hat i+\dfrac{1}{y^2}\hat j+\dfrac{1}{z^2}\hat k$
at $(1,1,1) \Rightarrow \vec{E}=\hat i+\hat j+\hat k$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electrostatic potential inside a charged spherical ball is given by $\phi=ar^2+b$, where r is the distance from the centre and a, b are constant. Then the charge density inside the ball is :

  1. $-6 a \epsilon _0r$

  2. $-24\pi a \epsilon _0r$

  3. $-6 a \epsilon _0$

  4. $-24 \pi a \epsilon _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Electric filed , $E=-\dfrac{d\phi}{dr}=-2ar$
By Gauss's law, $E.4\pi r^2=\dfrac{q _{in}}{\epsilon _0}$
$\Rightarrow q _{in}=(-2ar)4\pi r^2 \epsilon _0=-8\pi \epsilon _0 ar^3$
Now $\dfrac{dq _{in}}{dr}=-24\pi \epsilon _0 ar^2$ and $V=\dfrac{4}{3}\pi r^3,  \dfrac{dV}{dr}=4\pi r^2$
Charge density , $\rho=\dfrac{dq _{in}}{dV}=\dfrac{dq _{in}}{dr}\times \dfrac{dr}{dV}=(-24\pi \epsilon _0 ar^2)\times \dfrac{1}{4\pi r^2}=-6 \epsilon _0 a$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

An electric field is given by $\vec E = (y \hat i +  \hat x) NC^{-1}$. Find the work done (in $J$) by the electric field in moving a $1\ C$ charge from $\vec r _A = (2 \hat i + 2 j) m $ to $\vec r _B = (4 \hat i + \hat j) m$

  1. $0\ J$

  2. $-2\ J$

  3. $2\ J$

  4. $4\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done , $W=\int \vec{F}.\vec{dr}$

Here electrostatic force , $\vec{F}=q\vec{E}=q(y\hat i+x\hat j)$
$\vec{F}.\vec{dr}=q(y\hat i+x\hat j).(dx\hat i+dy\hat j)=q(ydx+xdy)=d(xy)$  as $q=1  C$

Now $W=\int _{2,2}^{4,1}d(xy)=[xy] _{2,2}^{4,1}=4\times 1-2\times 2=4-4=0$