Tag: relation between electric field and electric potential

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Questions Related to relation between electric field and electric 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 is represented by $E$, where $A=10\ V/{m}^{2}$. The electric potential at the origin with respect to the point $(10,20)m$ will be $V$ $(0,0)=.......\ volt$.

  1. $200$
  2. $300$
  3. $400$
  4. $500$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

E = -dV/dr. Given E = 10, V = -integral(E dr). Potential difference V(0,0) - V(10,20) = integral from 0 to 10 of E dx + integral from 0 to 20 of E dy. This requires more context on the field vector. Assuming a uniform field, the result is 500.

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

Electric potential is given by $V=6x-8{xy}^{2}$. Then electric force acting on $2\ C$ point charge placed at the origin will be

  1. $2\ N$
  2. $6\ N$
  3. $8\ N$
  4. $12\ N$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$V=6x-8xy^2$

$Q=2C$
$\begin{array}{l} { E _{ x } }=-\dfrac { { dv } }{ { dx } } =6-8 x { y^{ 2 } }=6 \ { E _{ y } }=-\dfrac { { dv } }{ { dy } } =-8\times x\times 2y=0 \ E=\sqrt { { E _{ x } }^{ 2 }+{ E _{ y } }^{ 2 } } =\sqrt { { { \left( 6 \right)  }^{ 2 } }+0 } =6 \ F=QE \ F=2\times 6=12N \end{array}$

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 electrostatic potential $V$ at any point (x, y, z) in space is given by $V = 4x^2$

  1. The y-and z-components of the electrostatic field at any point are zero.
  2. The x-component of electric field an any point is given by $(-8x \hat{i})$
  3. The x-component of electric field at $(1, 0, 2)$ is $(-8\hat{i})$

  4. The y-and z-components of the field are constant in magnitude.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation
We have $V = 4x^2$
So, the $x , y$ and $z$ components of the electrostatic field are

$E _x = \dfrac{-\partial V}{\partial x} = -8x$

$E _y = \dfrac{-\partial V}{\partial y} = 0$

$E _z = \dfrac{-\partial V}{\partial z} = 0$

So, $\overrightarrow{E} = E _x\hat{i} + E _y \hat{j} + E _z\hat{k} = -8x\hat{i}$. 
The electrostatic field at $(1, 0, 2)$ is $\overrightarrow{E} = (-8)\hat{i} \,V/m$.
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Two conducting shells of radii $2\ cm$ and $3\ cm$ are separately charged by $10\ V$ and $5\ V$ potential, respectively. Now smaller shell is placed inside bigger shell, and  then connected by a wire. What will be potential at the surface of smaller shell ?

  1. zero
  2. $\dfrac{35}{3}\ volt$
  3. $\dfrac{25}{3}\ volt$
  4. $\dfrac{10}{3}\ volt$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 on the x-axis electric potential decreases uniformly from 60 V to 20 V between x = -2 m to x = +2 m, then the magnitude of electric field at the origin

  1. Must be 10 V/m

  2. May be greater than 10 V/m

  3. Is zero

  4. Is 5 V/m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

E = -dV/dx. The potential changes by 40 V over 4 m. The average field is 10 V/m. Since it decreases uniformly, the field is constant at 10 V/m everywhere in that interval.

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

$64$ charged drops coalesce to form a bigger charged drop. The potential of bigger drop will be times that of smaller drop-

  1. $4$
  2. $16$
  3. $64$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume is conserved: 64 * (4/3)pi*r^3 = (4/3)pi*R^3, so R = 4r. Potential V = kQ/r. Q_new = 64q. V_new = k(64q)/(4r) = 16 * (kq/r) = 16V.

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 $10N/C$ exists in the vertically downward direction, the increase in the electric potential as one goes through a height of $50cm$ is:

  1. $20J$
  2. $\dfrac{1}{5}J$
  3. $5J$
  4. $\dfrac{1}{20}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Electric field $=10N/C$
Vertically downward direction electric potential as one goes through $h=50cm$ $=50\times { 10 }^{ -2 }m$
Now, $V=E/d$
$=10/50\times { 10 }^{ -2 }=\dfrac { 100 }{ 5 } =20J$
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 an electric field the potential at a point is given by the following relation $V = \dfrac{343}{r}$ where r is distance from the origin. The electric field at $r = 3\hat i + 2\hat j + 6\hat k $ is:

  1. $21\hat i + 14\hat j + 42\hat k $
  2. $3\hat i + 2\hat j + 6\hat k $
  3. $\dfrac{1}{7}(3\hat i + 2\hat j + 6\hat k )$
  4. $-(3\hat i + 2\hat j + 6\hat k )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

B. $3i+2j+6k$


Formula,

$E=\dfrac{V}{|\vec{r}|}\cdot \hat{r}$

$E=\dfrac{343}{|\vec{r}|^2}\cdot \dfrac{3i+2j+6k}{|r|}$

$=\dfrac{343}{7^2}\cdot \dfrac{3i+2j+6k}{7}$

$=3i+2j+6k$

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, the electric potential at a distance r?

  1. Is uniform in the region

  2. Is proportional to r
  3. Is proportional to $r^2$

  4. Increases as one goes away from the origin
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\quad E∝r\quad and\quad V=0\quad at\quad r=0$

$E=kr$
$E=\frac { -dv }{ dr } $
$V=-int{Edr}$
$V=-int { Krdr}$ 
$V=-k\frac { { r }^{ 2 } }{ 2 } +C$
$V=-k\frac { { r }^{ 2 } }{ 2 } $
$V=0\quad r=0\quad C=0$
$V=0\quad r=0\quad C=0$
 v is proportional to ${ r }^{ 2 }$