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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. <span>$ 2.5\times 10^{39} $</span>

  4. <span>$ 3 \times 10^{39} $</span>
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)
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

$V=6x-8xy^2$

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