Tag: relation between electric field and electric potential

Questions Related to 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)

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

  1. $V _E = -\int _C E.dl$

  2. $E _V = -\int _C V.dl$

  3. $V _E = -\int E.dl$

  4. $E _V = -\int E.dl$

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Correct Option: A
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:
$V=-\int _{C}^{ }E.dl$

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 $(1, -3, 2)$.

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

  1. $48 \hat j$

  2. $48 \hat i$

  3. $-48 \hat i$

  4. $-48 \hat j$

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Correct Option: A
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).$

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:

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

  1. $256\ V$

  2. $512\ V$

  3. $123\ V$

  4. $215\ V$

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Correct Option: A
Explanation:

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

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

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

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

  1. $-27$ $V$

  2. $-36$ $V$

  3. $27$ $V$

  4. $36$ $V$

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Correct Option: B
Explanation:
As we know, relation between uniform electric field strength $\&$ potential difference between two point a distance d
$V=-Ed \; \Rightarrow \; V=-(12)(3)=-36 \; Volts$

In the direction of electric field, the electric potential:

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

  1. decreases

  2. increases

  3. remains uncharged

  4. becomes zero

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Correct Option: A
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.

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

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

  1. $V\propto \displaystyle\frac{1}{r}$

  2. $V\propto r$

  3. $V\propto \displaystyle\frac{1}{r^2}$

  4. $V=$ constant

Show answer
Correct Option: D
Explanation:

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

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.

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 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 $\dfrac{V}{x _0}$;

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Correct Option: C

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:

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

  1. $2$ V/m

  2. $3\sqrt3$ V/m

  3. $2\sqrt2$V/m

  4. $5\sqrt3$ V/m

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Correct Option: B

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 :

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

  1. $5.5\times { 10 }^{ -13 }C$

  2. $7.5\times { 10 }^{ -13 }C$

  3. $4.5\times { 10 }^{ -12 }C$

  4. $2.5\times { 10 }^{ -11 }C$

Show answer
Correct Option: A
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 $

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,

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

  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 $V _x=\frac {\sigma}{\epsilon _0}x$

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Correct Option: A,B,D
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.