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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.

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

Which of the following is true for uniform electric field ?

  1. all points are at the same potential

  2. no two points can have the same potential

  3. pairs of points separated by the different distance must have the same difference in potential

  4. none of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

uniform electric field means the electric field vector does not vary with positive and electric field lines are parallel and equally spaced. the statement given define the equipotential surface, so answer is (d) -none of these.

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 and the electric potential at a point are E and V respectively. Then, the incorrect statements are :

  1. If E $=$ 0, V must be zero.

  2. If V $=$ 0, E must be zero.

  3. If E $\neq$ 0, V cannot be zero.

  4. If V $\neq$ 0, E cannot be zero.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

The electric field is $E=-\dfrac{dV}{dx}$
If $V=0$, we can not say $E$ must be zero, we say only $E$ may be zero.
If $V \neq 0 $, $E$ must be zero when $V$ is max i.e, $\dfrac{dV}{dx}=0$ For example, inside the conductor $E=0,$ but $V \neq 0 $
If $E \neq 0$ , $V$  may be zero when two equal and opposite charges separated by a distance and  at the midpoint in between the charges field is non-zero but potential is 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

A charge of $6.76$ $\mu$C in an electric field is acted upon by a force of $2.5 N$. The potential gradient at this point is :

  1. $3.71 \times 10^{15} Vm^{-1}$

  2. $-3.71 \times 10^{12} Vm^{-1}$

  3. $3.71 \times 10^{10} Vm^{-1}$

  4. $-3.71 \times 10^{5} Vm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$F=qE \Rightarrow E=\dfrac{F}{q}=\dfrac{2.5}{6.76\times 10^{-6}}=3.7\times 10^5$

The potential gradient is the electric field i.e, $E=-\nabla V=-3.7 \times 10^5 $

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 $120$ V to $80$ V as one moves on the X-axis from x $=$ -1 cm to x $=$ $+1$ cm. The electric field at the origin :

  1. must be equal to $20$ V/cm

  2. may be equal to $20$ V/cm

  3. may be greater than $20$ V/m

  4. may be less than $20$ V/cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the electric potential decreases uniformly so the electric field is uniform along the x axis.

$E=-\dfrac{dV}{dx}=\dfrac{120-80}{1-(-1)}=20 $V/m. It must be equal to 20 V/m at the origin.

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,

  1. it is uniform in the region

  2. it is proportional to r

  3. it is proportional to r$^2$

  4. it increases as one goes away from the origion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field is directly propotional to $r$, therefore
$E=kr$
We know that
$\ V=-\int _{ 0 }^{ r }{ \overrightarrow { E. } \overrightarrow { dr }  } $
which gives 
$V=-\dfrac { k{ r }^{ 2 } }{ 2 } $ since V at $r=0$ is $0$
Hence 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

A uniform electric field of $20$ NC$^{-1}$ exists along the x-axis in space. The potential difference V$ _B-$V$ _A$ for the point A $=$ $(4 m, 2m)$ and B $=$ $(6m, 5m)$ is:

  1. $20$ $\sqrt{13}$ V

  2. $- 40 V$

  3. zero V

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $\vec{E}=20 \hat i \Rightarrow E _x=20$
Potential difference ,$V _B-V _A=-\int _A^B Edr=-\int _4^6E _x dx=-20\int _4^6 dx=-20(6-4)=-40  V$

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 at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively. Out of the given points on the periphery of the circle, the potential is minimum at :

  1. A

  2. B

  3. C

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

The relation between electric field and potential is given by $\vec{E} = \displaystyle -\frac{\partial V}{\partial x}\hat{i} -\frac{\partial V}{\partial y}\hat{j} $

Given that, at origin, electric field is along positive x-axis.
Thus, $\displaystyle \frac{\partial V}{\partial x} < 0$ and $\displaystyle \frac{\partial V}{\partial y} = 0$
Thus, $V$ decreases in the positive x-direction and remains constant in y-direction.
Hence, minimum $V$ occurs at $(a,0)$ i.e., $A$