Tag: coulomb's law

Questions Related to coulomb's law

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.

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

  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 $\dfrac{V}{x _0}$;

Show answer
Correct Option: C

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:

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

  1. $2$ V/m

  2. $3\sqrt3$ V/m

  3. $2\sqrt2$V/m

  4. $5\sqrt3$ V/m

Show answer
Correct Option: B

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 :

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

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

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

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

  4. $2.5\times { 10 }^{ -11 }C$

Show answer
Correct Option: A
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 $

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,

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

  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 $V _x=\frac {\sigma}{\epsilon _0}x$

Show answer
Correct Option: A,B,D
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.

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 :

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

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

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

  3. zero

  4. $\cfrac{1}{\sqrt 3}(\hat { i } +\hat { j } +\check { k } )$

Show answer
Correct Option: B
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 $(1,1,1) \Rightarrow \vec{E}=\hat i+\hat j+\hat k$

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 :

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

  1. $-6 a \epsilon _0r$

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

  3. $-6 a \epsilon _0$

  4. $-24 \pi a \epsilon _0$

Show answer
Correct Option: C
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 , $\rho=\dfrac{dq _{in}}{dV}=\dfrac{dq _{in}}{dr}\times \dfrac{dr}{dV}=(-24\pi \epsilon _0 ar^2)\times \dfrac{1}{4\pi r^2}=-6 \epsilon _0 a$

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 $\vec r _B = (4 \hat i + \hat j) m$

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

  1. $0\ J$

  2. $-2\ J$

  3. $2\ J$

  4. $4\ J$

Show answer
Correct Option: A
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 $W=\int _{2,2}^{4,1}d(xy)=[xy] _{2,2}^{4,1}=4\times 1-2\times 2=4-4=0$

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 :

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

  1. $0$

  2. $-6\phi _0\varepsilon _0$

  3. $-2\phi _0\varepsilon _0$

  4. $\dfrac{-6\phi _0}{\varepsilon _0}$

Show answer
Correct Option: B
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 $

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:

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

  1. 3

  2. 4

  3. 5

  4. 6

Show answer
Correct Option: A
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$=|W|=3$

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

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

  1. $5.20\ N$

  2. $7.20\ N$

  3. $6.20\ N$

  4. $8.20\ N$

Show answer
Correct Option: B
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)