Tag: coulomb's law

Questions Related to coulomb's law

$64$ charged drops coalesce to form a bigger charged drop. The potential of bigger drop will be times that of smaller drop-

  1. $4$

  2. $16$

  3. $64$

  4. $8$


Correct Option: B

A uniform electric field $10N/C$ exists in the vertically downward direction, the increase in the electric potential as one goes through a height of $50cm$ is:

  1. $20J$

  2. $\dfrac{1}{5}J$

  3. $5J$

  4. $\dfrac{1}{20}J$


Correct Option: A
Explanation:
Electric field $=10N/C$
Vertically downward direction electric potential as one goes through $h=50cm$ $=50\times { 10 }^{ -2 }m$
Now, $V=E/d$
$=10/50\times { 10 }^{ -2 }=\dfrac { 100 }{ 5 } =20J$

In an electric field the potential at a point is given by the following relation $V = \dfrac{343}{r}$ where r is distance from the origin. The electric field at $r = 3\hat i + 2\hat j + 6\hat k $ is:

  1. $21\hat i + 14\hat j + 42\hat k $

  2. $3\hat i + 2\hat j + 6\hat k $

  3. $\dfrac{1}{7}(3\hat i + 2\hat j + 6\hat k )$

  4. $-(3\hat i + 2\hat j + 6\hat k )$


Correct Option: D
Explanation:

B. $3i+2j+6k$


Formula,

$E=\dfrac{V}{|\vec{r}|}\cdot \hat{r}$

$E=\dfrac{343}{|\vec{r}|^2}\cdot \dfrac{3i+2j+6k}{|r|}$

$=\dfrac{343}{7^2}\cdot \dfrac{3i+2j+6k}{7}$

$=3i+2j+6k$

The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero, the electric potential at a distance r?

  1. Is uniform in the region

  2. Is proportional to r

  3. Is proportional to $r^2$

  4. Increases as one goes away from the origin


Correct Option: C
Explanation:

$\quad E∝r\quad and\quad V=0\quad at\quad r=0$

$E=kr$
$E=\frac { -dv }{ dr } $
$V=-int{Edr}$
$V=-int { Krdr}$ 
$V=-k\frac { { r }^{ 2 } }{ 2 } +C$
$V=-k\frac { { r }^{ 2 } }{ 2 } $
$V=0\quad r=0\quad C=0$
$V=0\quad r=0\quad C=0$
 v is proportional to ${ r }^{ 2 }$

In a certain region of space, the potential is given by $V=k\left[ { 2x }^{ 2 }-{ y }^{ 2 }+{ z }^{ 2 } \right] $. The electric field at the point$ (1,1,1)$ has magnitude :

  1. $k\sqrt { 6 } $

  2. $2k\sqrt { 6 } $

  3. $2k\sqrt { 3 } $

  4. $4k\sqrt { 3 } $


Correct Option: B
Explanation:

Given, $V=k[2x^2-y^2+z^2]$

Electric field , $\vec{E}=-(\dfrac{dV}{dx}\hat i+\dfrac{dV}{dy}\hat j+\dfrac{dV}{dz}\hat {k})$

or,$\vec{E}=-k(4x\hat i-2y\hat j+2z\hat k)$

or,$\vec{E} _{(1,1,1)}=-k(4\hat i-2\hat j+2\hat k)$

Magnitude of electric field$ =|\vec{E} _{(1,1,1)}|=\sqrt{k^2(16+4+4)}=k\sqrt {24}=2k\sqrt 6$

Let V be electric potential and E the magnitude of the electric field. At a given position, which of the statement is true

  1. E is always zero where V is zero

  2. V is always zero where E is zero

  3. E can b zero where V is non zero

  4. E is always nonzero where V is nonzero


Correct Option: A

Two plates are 1 cm apart and the potential difference between them is 10 volt. The electric field between the plates is

  1. 10 N/C

  2. 250 N/C

  3. 500 N/C

  4. 1000 N/C


Correct Option: D

The equation of an equipotential line in an electric field is $y=2x$, then the electric field strength vector at $(1,2)$ may be :

  1. $4\hat { i } +3\hat { j } $

  2. $4\hat { i } +8\hat { j } $

  3. $8\hat { i } +4\hat { j } $

  4. $-8\hat { i } +4\hat { j } $


Correct Option: D
Explanation:

Now equation of equipotential surface is $y=2x$
Now electric field along the euipotential surface should be zero
therefore angle made by equipotential surface with x-axis is $tan^{-1} { (2) } $
Now since net electric field should be perpendicular to the equipotential surface
therefore for any electric field which makes an angle $tan^{-1} { (-1/2) } $ with x-axis can be the electric field at point $(1,2)$ which is true only for option (D)

because for two perpendicular line, product of their slope should be equal to -1 i.e., $m _1 \times m _2=-1$

Two plates are at potentials $-10 V$ and $+30 V$. If the separation between the plates is $2 cm$ then the electric field between them will be 

  1. 2000 V/m

  2. 1000 V/m

  3. 500 V/m

  4. 3000 V/m


Correct Option: A
Explanation:

Given,

$d=2cm$
$V _2-V _1=+30V-(-10)=40V$
Electric field, $E=\dfrac{V _2-V _1}{d}$
$E=\dfrac{40}{2\times 10^{-2}}=2000V/m$
The correct option is A.

In a certain region the electric potential at a point $(x, y, z)$ is given by the potential function $V = 2x + 3y - z$. Then the electric field in this region will :

  1. increase with increase in x and y

  2. increase with increase in y and z

  3. increase with increase in z and x

  4. remain constant


Correct Option: D
Explanation:

$V=2x+3y-z$

$E _x=-\dfrac{dV}{dx}=-2,  E _y=-\dfrac{dV}{dy}=-3 $ and $E _z=-\dfrac{dV}{dz}=1$

As the field components are independent of x,y and z so the field remains constant.