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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

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?

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

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

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

  4. <span>$x = -C/A, y = -2BC/A$ $^2$, any value of $z$</span>
Reveal answer Fill a bubble to check yourself
D Correct answer
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 . 

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 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 $(1 m, 1 m, 1 m)$ :

  1. <span>$20 \sqrt 2$ N/C</span>

  2. <span>$20 \sqrt 3$ N/C</span>

  3. <span>$10 \sqrt 3$ N/C</span>

  4. <span>$20 $ N/C</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

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 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 :

  1. $r^2$

  2. $r$

  3. $1/r$

  4. $1/r^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $r$.
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

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.

  1. I only

  2. II only

  3. I and II only

  4. I and III only

  5. None of the above
Reveal answer Fill a bubble to check yourself
E Correct answer
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.

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

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

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

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 (in $V/m$) is given by $E=10x^3$. Determine the potential difference, in volts, between $x=0m$ and $x=3m$.

  1. $202.5$

  2. $100$

  3. $20$

  4. $250$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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