Tag: coulomb's law

Quizzes (4) Questions (102)

Questions Related to coulomb's law

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
A Correct answer
Explanation

The electric potential decreases uniformly from V to -V over a distance of 2x_0 along the x-axis, meaning the magnitude of the uniform electric field is the total potential difference divided by the total distance. The total change in potential is 2V over a distance of 2x_0, giving an electric field magnitude of V/x_0. Since the question asks for the field at the origin and specifies uniform decrease, it must equal this value.

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 region, the potential is given by V=-{5x + 5y + 5z}, where V is in volts and x, y, z are in meters. The intensity of the electric field is:

  1. $2$ V/m
  2. $3\sqrt3$ V/m
  3. $2\sqrt2$V/m
  4. $5\sqrt3$ V/m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field intensity is the negative gradient of the potential V. Taking partial derivatives, Ex = -dV/dx = 5, Ey = -dV/dy = 5, and Ez = -dV/dz = 5. The magnitude of the electric field is the square root of the sum of the squares of these components, which is sqrt(5^2 + 5^2 + 5^2) = sqrt(75) = 5*sqrt(3) V/m.

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

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

If the electrostatic potential is given by $\phi =\phi _0(x^2+ y^2 + z^2)$ where $\phi _0$ is constant, then the charge density of the given potential would be :

  1. $0$
  2. $-6\phi _0\varepsilon _0$
  3. $-2\phi _0\varepsilon _0$
  4. $\dfrac{-6\phi _0}{\varepsilon _0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ \overrightarrow{E} = -\triangledown \phi $
$ \overrightarrow{\triangledown}.\overrightarrow{E} = \rho/\epsilon _0 $
Now, $\phi = \phi _0 (x^2 + y^2 + z^2) \Rightarrow \overrightarrow{E} = -2\phi _0 ( \hat{i}+\hat{j}+\hat{k} ) \Rightarrow \rho = -6\phi _0 \epsilon _0 $

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 field in a region is given as $\bar{E}=x\hat{i}+2y\hat{j}+3\hat{k}$. In this region point A(3,3,1) and point B (4,2,1) are there. The magnitude of work done by the electric field, if 2 coulomb charge is moved from A to B. All values are in SI units:

  1. 3

  2. 4

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $\vec{E}=x\hat{i}+2y\hat{j}+3\hat{k}$ and $ q=2 C$
Work done, $W=\int^B _Aq\vec{E}.\vec{dr}=q\int^B _A(x\hat{i}+2y\hat{j}+3\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$
or,$W=2\int^{(4,2,1)} _{(3,3,1)}xdx+2ydy+3dz=2[\frac{16-9}{2}+(4-9)+3(1-1)]=7-10=-3$
Magnitude of work done$=|W|=3$

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

Find the magnitude of the force on a charge of $12\mu C$ placed at point where the potential gradient has a magnitude of $6\times 10^{5}V\ m^{-1}$

  1. $5.20\ N$
  2. $7.20\ N$
  3. $6.20\ N$
  4. $8.20\ N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potential gradient is nothing but the rate of change of electric potential with position and it is equal to electric field at that point.


$\dfrac{dV}{dl}=E$=electric field

$\implies E=6\times 10^5Vm^{-1}$

Force on charge $=F=qE=12\times 10^{-6}\times 6\times 10^5$

$\implies F=7.2N$

Answer-(B)