Tag: physics

Questions Related to physics

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. At which points is the electric field equal to zero?

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

  1. $x = +C/A, y = +BC/A$ $^2$, any value of $z$

  2. $x = +C/A, y = +2BC/A$ $^2$, any value of $z$

  3. $x = -C/A, y = -2BC/A$ $^2$, $z=0$

  4. $x = -C/A, y = -2BC/A$ $^2$, any value of $z$

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Correct Option: D
Explanation:

$ \vec E = - \triangledown V = - [ ( Ay - 2Bx) \hat i + ( Ax + C) \hat j ] $


$ \vec E = 0$  at, 

$ E _y = 0 \Rightarrow Ax + C= 0 \Rightarrow x = -C/A $

$ E _x= 0 \Rightarrow Ay - 2Bx= Ay + 2BC/A =0 \Rightarrow y = -2BC/A^2 $

$ E _z = 0 $  everywhere . 

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. If A is $10$ SI units, find the magnitude of the electric field at $(1 m, 1 m, 1 m)$ :

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

  1. $20 \sqrt 2$ N/C

  2. $20 \sqrt 3$ N/C

  3. $10 \sqrt 3$ N/C

  4. $20 $ N/C

Show answer
Correct Option: B
Explanation:

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] = -10[2 \hat i + 2 \hat j + 2 \hat k ] = 20\sqrt{3} N/C$

The electric potential at a point (x, y) in the x-y plane is given by V = - Kxy. The field intensity at a distance r in this plane, from the origin is proportional to :

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

  1. $r^2$

  2. $r$

  3. $1/r$

  4. $1/r^2$

Show answer
Correct Option: B
Explanation:
Let the $x-y$ coordinates of the point at distance $r$ from the origin be given as $x=rcos\theta$, $y=rsin\theta$
Potential is given as $V=-Kxy$
Now, Electric field intensity is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=K(y\vec i+x\vec j)=K(rsin\theta\vec i+rcos\theta\vec j)=Kr(sin\theta\vec i+cos\theta\vec j)\propto r$
So electric field potential is proportional to $r$.

Find out the relationship between the electric field and electric potential include which of the following statement?

I. If the electric field at a certain point is zero, then the electric potential at the same point is also zero.

II. The electric potential is inversely proportional to the strength of the electric field.

III. If the electric potential at a certain point is zero, then the electric field at the same point is also zero.

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

  1. I only

  2. II only

  3. I and II only

  4. I and III only

  5. None of the above

Show answer
Correct Option: E
Explanation:

If electric field at some point is zero, it is not necessary for the electric potential to be the same. Consider the mid point of line joining to equal charges of same sign. Field there is zero, but potential is finite and positive.


Electric field strength is given by $E=\dfrac{kQq}{r^2}$
Electric potential is given by $V=\dfrac{dQq}{r}$
Clearly they are not inversely proportional to each other.

If electric potential at some point is zero, it is not necessary for the electric field to be the same. Consider the mid point of line joining to equal charges of opposite signs. Potential there is zero, but electric field exists.

When negative charges are kept in electric field then negative charges are accelerated by electric fields toward points:

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

  1. at lower electric potential

  2. at higher electric potential

  3. where the electric field is zero

  4. where the electric field is weaker

  5. where the electric field is stronger

Show answer
Correct Option: B
Explanation:

We know that , $F=-\Delta U$, change in potential energy. 

As charge is negative so, $F=-qE=-\Delta U$ or $\Delta U=qE$
So we will get change in potential as positive and hence the electric field should be towards  points at higher potential.

An electric field (in $V/m$) is given by $E=10x^3$. 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. $202.5$

  2. $100$

  3. $20$

  4. $250$

Show answer
Correct Option: A
Explanation:

We know that, $E=-\dfrac{dV}{dx}$


$V=-\int _0^3 Edx=-\int _0^3 10x^3 dx=-10\times \dfrac{3^4}{4}=202.5 $

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 unchanged

  4. Becomes zero

Show answer
Correct Option: A
Explanation:

$dV=-E.dr$, so the electric potential $V$ decreases continuously as we move along the direction of the electric field

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

Show answer
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. $-48 \hat j$

Show answer
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. $215\ V$

Show answer
Correct Option: A
Explanation:

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

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