Tag: field strength and potential gradient

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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 electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,

  1. it is uniform in the region

  2. it is proportional to r
  3. it is proportional to r$^2$

  4. it increases as one goes away from the origion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field is directly propotional to $r$, therefore
$E=kr$
We know that
$\ V=-\int _{ 0 }^{ r }{ \overrightarrow { E. } \overrightarrow { dr }  } $
which gives 
$V=-\dfrac { k{ r }^{ 2 } }{ 2 } $ since V at $r=0$ is $0$
Hence V is proportional to ${ r }^{ 2 }$

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 $20$ NC$^{-1}$ exists along the x-axis in space. The potential difference V$ _B-$V$ _A$ for the point A $=$ $(4 m, 2m)$ and B $=$ $(6m, 5m)$ is:

  1. $20$ $\sqrt{13}$ V
  2. $- 40 V$

  3. zero V

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $\vec{E}=20 \hat i \Rightarrow E _x=20$
Potential difference ,$V _B-V _A=-\int _A^B Edr=-\int _4^6E _x dx=-20\int _4^6 dx=-20(6-4)=-40  V$

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

The electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively. Out of the given points on the periphery of the circle, the potential is minimum at :

  1. A

  2. B

  3. C

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

The relation between electric field and potential is given by $\vec{E} = \displaystyle -\frac{\partial V}{\partial x}\hat{i} -\frac{\partial V}{\partial y}\hat{j} $

Given that, at origin, electric field is along positive x-axis.
Thus, $\displaystyle \frac{\partial V}{\partial x} < 0$ and $\displaystyle \frac{\partial V}{\partial y} = 0$
Thus, $V$ decreases in the positive x-direction and remains constant in y-direction.
Hence, minimum $V$ occurs at $(a,0)$ i.e., $A$

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

It is found that air breaks down electrically, when the electric field is $  3 \times 10^{6} \mathrm{V} / \mathrm{m} .  $ What is the potential to which a sphere of radius $1  \mathrm{m}  $ can be raised, before sparking takes place?

  1. $ V=10^{6} \mathrm{V} $
  2. $ V=2 \times 10^{6} \mathrm{V} $
  3. $ V=3 \times 10^{6} \mathrm{V} $
  4. $ V=4 \times 10^{6} \mathrm{V} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a charged conducting sphere, the electric field at the surface is E = kQ/R^2 and the potential is V = kQ/R. Therefore, V = E * R. Given E = 3 * 10^6 V/m and R = 1 m, V = 3 * 10^6 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

In moving from A to B along an electric field line, the wok done by the electric field on an electron is $6.4 \times 10^{-19}$ J. If $\phi _1$ and $\phi _2$ are equipotential surfaces, then the potential difference $V _b-V _A $ is

  1. -4V

  2. 4V

  3. zero

  4. 6.4 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done by the electric field is W = q * (V_A - V_B). For an electron, q = -1.6 * 10^-19 C. Given W = 6.4 * 10^-19 J, we have 6.4 * 10^-19 = -1.6 * 10^-19 * (V_A - V_B), which simplifies to V_A - V_B = -4 V, or V_B - V_A = 4 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

The equation of an equipotential line is an electric field is y = 2x, then the electric field strength vector at (1, 2) may be 

  1. $4\vec{i} + 3\vec{j}$
  2. $4\vec{i} + 8\vec{j}$
  3. $8\vec{i} + 4\vec{j}$
  4. $-8\vec{i} + 4\vec{j}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field vector is always perpendicular to the equipotential line at any given point. The slope of the equipotential line y = 2x is m1 = 2. Therefore, the slope of the electric field vector must be the negative reciprocal, m2 = -1/2. Testing the options for a vector (E_x, E_y) with slope -1/2 = E_y / E_x: option D gives E_x = -8 and E_y = 4, so E_y / E_x = 4 / (-8) = -1/2, which matches correctly.

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 in a certain region along the x-axis varies with x according to the relation $V(x) = 5 - 4x^2$. Then, the correct statement is :

  1. the potential difference between the points $x =1$m and $x=2$m is $12$ Volt
  2. the force experienced by a Coulomb of charge placed at $x =1$ m is $8$ Newton

  3. the electric field components along Y and Z direction are zero

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$V(x)=5-4x^2$

$V(1)=5-4=1 V,  V(2)=5-4(2^2)=-11 V$

Potential difference between $x=1 m$ and $x=2 m$ is $V _{12}=V _1-V _2=1-(-11)=12 V$

here, $E _x=-\dfrac{dV}{dx}=8x,  E _y=-\dfrac{dV}{dy}=0$ and $E _z=-\dfrac{dV}{dz}=0$

The electric force on $1$ coulomb charge at $x=1$ is $F=qE _x=1(8)=8 N$

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 point charge q moves from point P to a point S along a path PQRS in a uniform electric field E pointing parallel to the x-axis. The coordinates of P, Q. R and S are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$. The work done by the field in the above process is :

  1. $zero$
  2. $qEB$
  3. $qEa$
  4. $-qEa$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the field E is uniform, so E is constant at every point.
As E is directed parallel to x axis, so $\vec{E}=E\hat i$
The work done , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

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 potential is given by : $V = k {[2x^2 - y^2 + z^2]}$. The electric field at the point (1, 1, 1) has magnitude = 

  1. $k\sqrt{6}$
  2. $2k\sqrt{6}$
  3. $2k\sqrt{3}$
  4. $4k\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field is E = -grad(V). E = -(dV/dx i + dV/dy j + dV/dz k). Given V = k(2x^2 - y^2 + z^2), E = -k(4x i - 2y j + 2z k). At (1, 1, 1), E = -k(4 i - 2 j + 2 k). The magnitude is k * sqrt(4^2 + (-2)^2 + 2^2) = k * sqrt(16 + 4 + 4) = k * sqrt(24) = 2k * sqrt(6).

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 3C moving in a uniform electric field experiences a force of $3000 N$. The potential difference between two points situated in the field at a distance $1 cm$ from each other will be

  1. $10 V$
  2. $90 V$
  3. $1000 V$
  4. $9000 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field E = F/q = 3000 N / 3 C = 1000 N/C. The potential difference V = E * d. With d = 1 cm = 0.01 m, V = 1000 * 0.01 = 10 V.