Tag: electric field as gradient of potential

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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

The most appropriate relationship between electric field and electric potential can be described as 
($C$ is an arbitrary path connecting the point with zero potential infinity)

  1. $V _E = -\int _C E.dl$
  2. $E _V = -\int _C V.dl$
  3. $V _E = -\int E.dl$
  4. $E _V = -\int E.dl$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $\Delta V=V _{B}-V _{A}$ be the electrostatic potential energy difference between any two points A and B in the electric field. The electric potential difference between points A and B is given by:
$\Delta V= V _{B}-V _{A}=\dfrac{\Delta V}{q _{0}}=\dfrac{V _{B}-V _{A}}{q _{0}}$
$\Delta V=V _{B}-V _{A}=-q _{0}\int _{A}^{B}\overrightarrow{E}.\overrightarrow{dl}$
$\Delta V=\dfrac{\Delta V}{q _{0}}=-\int _{A}^{B}\overrightarrow{E}.\overrightarrow{dl}$      ...(i)
If C is an arbitrary path connecting the point with zero potential at infinity. Then, equation (i) becomes:
$V=-\int _{C}^{ }E.dl$
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 potential in a certain region of space is given by the function $xy^2z^3$ with respect to some reference point. Find the y-component of the electric field at $(1, -3, 2)$.

  1. $48 \hat j$
  2. $48 \hat i$
  3. $-48 \hat i$
  4. $-48 \hat j$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

given potential $=x{ y }^{ 2 }{ z }^{ 2 }$

we have to find the y-component of electric field at $(1,-3,2)$
here, we use the relation b/w electric field and the potential $E=\cfrac { -dv }{ dr } $
To find y- c ordinate of electric field, we differential function of V and y.
so,${ E } _{ y }=\cfrac { -dv }{ dy } =\cfrac { -d }{ dy } (x{ y }^{ 2 }{ z }^{ 3 })$
${ E } _{ y }=-2x{ y }{ z }^{ 3 }$
To find electric field at pt. $(1,-3,2)$
we substitute for $x=1$
$y=-3$
$z=2$
we get${ E } _{ y }=-2\left( 1 \right) \left( -3 \right) { \left( 2 \right)  }^{ 3 }$
$=48$
Hence the answer is $48\hat {j}$
so, the correct answer is option $ (a).$

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 $4\times 10^{20}eV$ of energy is required to move a charge of $0.25$ coulomb between two points, the p.d between them is:

  1. $256\ V$
  2. $512\ V$
  3. $123\ V$
  4. $215\ V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$4\times 10^{20}eV=4\times 10^{20}\times 1.6\times 10^{-19}=64 J$

So $E=64=V\times Q=0.25V$
$V=256 V$

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 electric field  of $12$ $V/m$ is along the positive $x$ direction. Determine the potential difference in volts, between $x=0m$ and $x=3m$.

  1. $-27$ $V$
  2. $-36$ $V$
  3. $27$ $V$
  4. $36$ $V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
As we know, relation between uniform electric field strength $\&$ potential difference between two point a distance d
$V=-Ed \; \Rightarrow \; V=-(12)(3)=-36 \; Volts$
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 the direction of electric field, the electric potential:

  1. decreases

  2. increases

  3. remains uncharged

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

In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of electric field.

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

Variation of potential V with distance r in electric field of E$=0$ is?

  1. $V\propto \displaystyle\frac{1}{r}$
  2. $V\propto r$
  3. $V\propto \displaystyle\frac{1}{r^2}$
  4. $V=$ constant
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Potential difference between two points is given by    $\Delta V = -E.r$
Given :  $E = 0$
$\implies  \ \Delta V = 0$
$\implies \ V =$ constant

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 decreases uniformly from V to -V along X-axis in a coordinate system as we moves from a (-$x _0$, 0) to ($x _0$, 0), then the electric field at the origin.

  1. must be equal to $\dfrac{V}{x _0}$;
  2. may be equal to $\dfrac{V}{x _0}$;
  3. must be greater than $\dfrac{V}{x _0}$;
  4. may be less than $\dfrac{V}{x _0}$;
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric potential decreases uniformly from V to -V over a distance of 2x_0 along the x-axis, meaning the magnitude of the uniform electric field is the total potential difference divided by the total distance. The total change in potential is 2V over a distance of 2x_0, giving an electric field magnitude of V/x_0. Since the question asks for the field at the origin and specifies uniform decrease, it must equal this value.

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 region, the potential is given by V=-{5x + 5y + 5z}, where V is in volts and x, y, z are in meters. The intensity of the electric field is:

  1. $2$ V/m
  2. $3\sqrt3$ V/m
  3. $2\sqrt2$V/m
  4. $5\sqrt3$ V/m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field intensity is the negative gradient of the potential V. Taking partial derivatives, Ex = -dV/dx = 5, Ey = -dV/dy = 5, and Ez = -dV/dz = 5. The magnitude of the electric field is the square root of the sum of the squares of these components, which is sqrt(5^2 + 5^2 + 5^2) = sqrt(75) = 5*sqrt(3) 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

A copper ball of radius 1 cm work function 4.47 eV is irradiated with ultraviolet radiation of wavelength $2500\mathring { A } $. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be finite value of the potential on the ball. The charge acquired by the ball is :

  1. $5.5\times { 10 }^{ -13 }C$
  2. $7.5\times { 10 }^{ -13 }C$
  3. $4.5\times { 10 }^{ -12 }C$
  4. $2.5\times { 10 }^{ -11 }C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From photo electric effect equation :

]
$h\nu=h\nu _{0} + K.E _{max}$


so Maximum kinetic energy will be


$K.E _{max}= \dfrac{hc}{\lambda} - h\nu _{0}$
 
putting the given values in the above equation

$K.E _{max} = e\times V$ 

so V will be 

$V= \dfrac{k\times Q}{r}$
 
:: $ q = 5.5\times 10^{-13} C $

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 infinite, parallel, non-conducting sheets carry equal positive charge density $\sigma$. One is placed in the yz plane at $x=0$ and the other at distance $x=a$. Take potential $V=0$ at $x=0$. Then,

  1. for $0\leq x \leq a$, potential $V _x=0$
  2. for $x\geq a$, potential $V _x=-\frac {\sigma}{\epsilon _0}(x-a)$
  3. for $x\geq a$, potential $V _x=\frac {\sigma}{\epsilon _0}(x-a)$
  4. for $x\leq 0$ potential $V _x=\frac {\sigma}{\epsilon _0}x$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Now , Since both are infinite plates and carry same charge density therefore, electric field between them will be equal to zero.
Now, potential will be constant between them and at $x=0; V=0$ and V = constant between the plates.
Therefore, V=0 between the plates means $0\le x\le a$
Now electric field beyond $x=a$ is $2\times \sigma/2\epsilon _o=\sigma/\epsilon _o$
We know that,
$V=-\int _{ a }^{ x }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=-E(x-a)$
$V=-\sigma(x-a)/\epsilon _o$

and for $x<0$
$V=-\int _{ x }^{ 0 }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=Ex$
$V=\sigma x/\epsilon _o$
option (A)(B)(D) are correct.