Tag: physics

Questions Related to physics

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

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

  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

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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:

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

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

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

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

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

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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)

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

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

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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?

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

  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

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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

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

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

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

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

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

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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 :

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

  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.

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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:

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. $4V$

  3. $8V$

  4. $10V$

Show answer
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)

An electric field is represented by $E$, where $A=10\ V/{m}^{2}$. The electric potential at the origin with respect to the point $(10,20)m$ will be $V$ $(0,0)=.......\ volt$.

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

  1. $200$

  2. $300$

  3. $400$

  4. $500$

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Correct Option: D

A force of 3000 N is acting on a charge of 4 coloumb moving in a uniform electric field. The potential difference between two point at a distance of 1 cm in this 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. 10 V

  2. 90 V

  3. 750 V

  4. 9000 V

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Correct Option: C

Electric potential is given by $V=6x-8{xy}^{2}$. Then electric force acting on $2\ C$ point charge placed at the origin will be

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

  1. $2\ N$

  2. $6\ N$

  3. $8\ N$

  4. $12\ N$

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Correct Option: D
Explanation:

$V=6x-8xy^2$

$Q=2C$
$\begin{array}{l} { E _{ x } }=-\dfrac { { dv } }{ { dx } } =6-8 x { y^{ 2 } }=6 \ { E _{ y } }=-\dfrac { { dv } }{ { dy } } =-8\times x\times 2y=0 \ E=\sqrt { { E _{ x } }^{ 2 }+{ E _{ y } }^{ 2 } } =\sqrt { { { \left( 6 \right)  }^{ 2 } }+0 } =6 \ F=QE \ F=2\times 6=12N \end{array}$