Tag: physics

Questions Related to physics

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} $


Correct Option: C

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


Correct Option: B

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. $-8\vec{i} + 4\vec{j}$


Correct Option: B

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


Correct Option: D
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 $F=qE _x=1(8)=8 N$

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. $-qEa$


Correct Option: D
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 , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

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. $4k\sqrt{3}$


Correct Option: B

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. $9000 V$


Correct Option: A

The potential at a point $x$ (measured in $\mu m )$ due to somecharges situated on the $x$ -axis is given by $V ( x ) = 20 / \left( x ^ { 2 } - 4 \right)$Volts. The electric field $E$ at $x = 4 \mu m$ is given by

  1. 5$/ 3$ Volt / \mum and in the -ve $x$ direction

  2. 5$/ 3$ Volt $/ \mu m$ and in the +ve $x$ direction

  3. 10$/ 9$ Volt / \mum and in the -ve $x$ direction

  4. 10$/ 9$ Volt $/ \mu m$ and in the +ve $x$ direction


Correct Option: B

Variation in potential is maximum if one goes :

  1. along the line of force

  2. perpendicular to the line of force

  3. in any direction

  4. none of these


Correct Option: A
Explanation:

$dV=-\vec{E}.d\vec{r}=-Edr cos\theta$


Hence, variation will be maximimum for $\theta=0^{o}$ or $180^{o}$, that is variation $dV$ is maximum along line of field or say line of force.

Answer-(A)

The electric field lines are closer together near object $A$ than they are near object $B$. We can conclude that :

  1. the potential near $A$ is greater than the potential near $B$

  2. the potential near $A$ is less than the potential near $B$

  3. the potential near $A$ is equal to the potential near $B$

  4. nothing about the relative potentials near $A$ and $B$


Correct Option: D
Explanation:

Potential decreases in the direction of electric field. So it depends  on whether the lines of forces are from $A$ to $B$ or from $B$ to $A$.