Tag: relation between electric field and electric potential

Questions Related to relation between electric field and electric potential

If we move in a direction opposite to the electric lines of force:

  1. electrical potential decreases

  2. electrical potential increases

  3. electrical potential remains uncharged

  4. nothing can be said.


Correct Option: B

A uniform wire $10 \,cm$ long is carrying a steady current. The potential drop across it is $10V$. The electric field inside it is _____

  1. zero

  2. $1Nm^{-1}$

  3. $10 \,Vm^{-1}$

  4. $100 \,Vm^{-1}$


Correct Option: D
Explanation:

$E = \dfrac{V}{r} = \dfrac{10V}{10 \times 10^{-2}m} = \dfrac{10 \times 100}{10} = 100 \dfrac{v}{m}$

The electric potential while moving along the lines of force

  1. decreases

  2. increases

  3. remains constant

  4. becomes infinite


Correct Option: A

$E=-\dfrac{dV}{dr}$, here negative sign signified that

  1. E is opposite to V

  2. E is negative

  3. E increases when V decreases

  4. E is directed in the direction of decreasing V


Correct Option: D
Explanation:

The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.

The ratio of electric force $ ( F _e ) $ to gravitational force acting between two electrons will be:

  1. $ 1 \times 10^{36} $

  2. $ 2 \times 10^{39} $

  3. $ 2.5\times 10^{39} $

  4. $ 3 \times 10^{39} $


Correct Option: C

Electric potential at ( x, y, z ) is given as $V$= $- x ^ { 2 } y \sqrt { z }$ Find the electrical field at (2 ,1, 1)

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

  2. $- 4 \hat { i } - 4 \hat { j } - 2 \hat { k }$

  3. $- 4 \hat { i } - 4 \hat { j } - 4 \hat { k }$

  4. $4 \hat { 1 } + 4 \hat { j } + 2 \hat { k }$


Correct Option: D

The electric field and the electric potential at a point inside a shell are E and V respectively. Which of the following is correct?

  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. None of these


Correct Option: D
Explanation:

In a shell $E=0 $ but $V \neq 0$.
Along the equatorial line of a dipole, $V=0 $ but $E  \neq 0$.

A charge of $6.25\mu C$ in an electric field is acted upon by a force $2.5N$. The potential gradient at this point is

  1. $4\times 10^{5}V / m$

  2. $4\times 10^{6}V / m$

  3. $2.5\times 10^{-6}V / m$

  4. $4\times 10^{7}V / m$


Correct Option: A
Explanation:

given force$ = 2.5$ $\mu$
we know $F = Eq$
$\Rightarrow 2.5=E( 6.25\ \mu c)$ 

$\Rightarrow E=\dfrac{2.5}{6.25\times 10^{-6}}$

$\Rightarrow E=4\times 10^5\ V/m$
$\therefore\ Potential\ gradient\ = \dfrac{dV}{dx}=E = 4\times 10^5\ V/m$

Electric potential $V$ at some point in space is zero. Then at that point :

  1. Electric intensity is necessarily zero.

  2. Electric intensity is necessarily non zero.

  3. Electric intensity may or may not be zero.

  4. Electric intensity is necessarily infinite.


Correct Option: C
Explanation:
$ E=-\dfrac { dV }{ dl } $ 
where $E=$ Electric field intensity ;  $V=$ Electric Potential ; $l=$ distance traveled in direction of electric field    
Electric intensity at a point is the negative of rate of change of the electric potential at a point. So if a function is zero at a given point, the slope will not necessarily be zero. 

In a uniform electric field, the potential is $10V$ at the origin of coordinates, and $8V$ at each of the points $(1,0,0),(0,1,0)$ and $(0,0,1)$. The potential at the point $(1,1,1)$ will be:

  1. $0$

  2. $4V$

  3. $8V$

  4. $10V$


Correct Option: B
Explanation:

The field is uniform. Hence ðv/ðr=constant=p(let)

Hence v=V°+p(i+j+k)

Now at origin v=V°=10volt

Also at the 3 points the value of the voltage are 8volt each. Hence p= -2

At (1,1,1) The voltage is=10 -2(1+1+1)

                                 =4 volt(ans)