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

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

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

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 $6.25\mu C$ 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. $4\times 10^{7}V / m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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.
Reveal answer Fill a bubble to check yourself
C Correct answer
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. 
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 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. $10V$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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 $V$ $(0,0)=.......\ volt$.

  1. $200$
  2. $300$
  3. $400$
  4. $500$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

E = -dV/dr. Given E = 10, V = -integral(E dr). Potential difference V(0,0) - V(10,20) = integral from 0 to 10 of E dx + integral from 0 to 20 of E dy. This requires more context on the field vector. Assuming a uniform field, the result is 500.

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

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

  1. $2\ N$
  2. $6\ N$
  3. $8\ N$
  4. $12\ N$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$V=6x-8xy^2$

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

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

  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.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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 $\overrightarrow{E} = (-8)\hat{i} \,V/m$.
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 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 ?

  1. zero
  2. $\dfrac{35}{3}\ volt$
  3. $\dfrac{25}{3}\ volt$
  4. $\dfrac{10}{3}\ volt$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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

  1. Must be 10 V/m

  2. May be greater than 10 V/m

  3. Is zero

  4. Is 5 V/m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

E = -dV/dx. The potential changes by 40 V over 4 m. The average field is 10 V/m. Since it decreases uniformly, the field is constant at 10 V/m everywhere in that interval.