Tag: coulomb's law

Questions Related to coulomb's law

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. $10V$

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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:

$V=6x-8xy^2$

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

The electrostatic potential $V$ at any point (x, y, z) in space is given by $V = 4x^2$

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

  1. The y-and z-components of the electrostatic field at any point are zero.

  2. The x-component of electric field an any point is given by $(-8x \hat{i})$

  3. The x-component  of electric field at $(1, 0, 2)$ is $(-8\hat{i})$

  4. The y-and z-components of the field are constant in magnitude.

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Correct Option: A,B,C,D
Explanation:
We have $V = 4x^2$
So, the $x , y$ and $z$ components of the electrostatic field are

$E _x = \dfrac{-\partial V}{\partial x} = -8x$

$E _y = \dfrac{-\partial V}{\partial y} = 0$

$E _z = \dfrac{-\partial V}{\partial z} = 0$

So, $\overrightarrow{E} = E _x\hat{i} + E _y \hat{j} + E _z\hat{k} = -8x\hat{i}$. 
The electrostatic field at $(1, 0, 2)$ is $\overrightarrow{E} = (-8)\hat{i} \,V/m$.

Two conducting shells of radii $2\ cm$ and $3\ cm$ are separately charged by $10\ V$ and $5\ V$ potential, respectively. Now smaller shell is placed inside bigger shell, and  then connected by a wire. What will be potential at the surface of smaller shell ?

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

  1. zero

  2. $\dfrac{35}{3}\ volt$

  3. $\dfrac{25}{3}\ volt$

  4. $\dfrac{10}{3}\ volt$

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Correct Option: C

If on the x-axis electric potential decreases uniformly from 60 V to 20 V between x = -2 m to x = +2 m, then the magnitude of electric field at the origin

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

  1. Must be 10 V/m

  2. May be greater than 10 V/m

  3. Is zero

  4. Is 5 V/m

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Correct Option: A