Tag: field strength and potential gradient

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Questions Related to field strength and potential gradient

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 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.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electric field lines point from higher to lower potential. Moving in the direction opposite to the lines of force means moving toward higher potential.

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 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. $100 \,Vm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

The electric potential while moving along the lines of force

  1. decreases

  2. increases

  3. remains constant

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

Electric field lines are defined as the path along which a positive test charge would move. Since a positive charge moves from higher to lower potential, the potential decreases along the lines of force.

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

$E=-\dfrac{dV}{dr}$, 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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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 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} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of electrostatic force to gravitational force between two electrons is approximately 4.17 * 10^42. The provided options are all in the 10^39 range, which is a common textbook approximation for this ratio.

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 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. $4 \hat { 1 } + 4 \hat { j } + 2 \hat { k }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field is the negative gradient of potential: E = -∇V. Computing partial derivatives: ∂V/∂x = -2xy√z, ∂V/∂y = -x²√z, ∂V/∂z = -x²y/(2√z). At point (2,1,1): Ex = -2(2)(1)(1) = -4, Ey = -(4)(1) = -4, Ez = -(4)(1)/(2×1) = -2. Therefore E = -(-4i - 4j - 2k) = 4i + 4j + 2k. The key is applying the gradient operator correctly and evaluating at the given point. Note that option D has a typo (should be î, not 1̂) but is clearly the intended 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

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)