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

$64$ 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. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume is conserved: 64 * (4/3)pi*r^3 = (4/3)pi*R^3, so R = 4r. Potential V = kQ/r. Q_new = 64q. V_new = k(64q)/(4r) = 16 * (kq/r) = 16V.

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 $10N/C$ 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. $\dfrac{1}{20}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Electric field $=10N/C$
Vertically downward direction electric potential as one goes through $h=50cm$ $=50\times { 10 }^{ -2 }m$
Now, $V=E/d$
$=10/50\times { 10 }^{ -2 }=\dfrac { 100 }{ 5 } =20J$
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 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. $-(3\hat i + 2\hat j + 6\hat k )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


Formula,

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

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

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

$=3i+2j+6k$

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 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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 ${ r }^{ 2 }$

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 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 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field and electric potential are related by E = -grad(V). A region can have zero electric field while having a non-zero constant electric potential (such as inside a charged conducting sphere). Thus, E can be zero where V is non-zero.

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 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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field E between two parallel plates is given by E = V/d. Here, V = 10 V and d = 1 cm = 0.01 m, so E = 10 / 0.01 = 1000 N/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

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 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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., $m _1 \times m _2=-1$

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 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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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 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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.