Question 11 Mark
Write the value of the electric field due to an electric dipole at a point situated on its axis.
Answer
View full question & answer→$E=\frac{1}{4 \pi \epsilon_0} \times \frac{2 p}{r^3}$
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Question 21 Mark
Write the formula to determine the value of electric field intensity due to a uniformly charged thin spherical shell.
Answer
View full question & answer→(i) Electric field outside shell: $E =\frac{q}{4 \pi \epsilon_0 r^2}$
Here, $q=4 \pi R^2 \sigma$ is the total charge on spherical shell
Vector form $\vec{E}=\frac{q}{4 \pi \epsilon_0 r^2} \vec{r}$
(ii) Electric field inside the shell : Due to a uniformly charged thin spherical shell, the electric field at all points inside it is zero.
Here, $q=4 \pi R^2 \sigma$ is the total charge on spherical shell
Vector form $\vec{E}=\frac{q}{4 \pi \epsilon_0 r^2} \vec{r}$
(ii) Electric field inside the shell : Due to a uniformly charged thin spherical shell, the electric field at all points inside it is zero.
Question 31 Mark
Draw lines of force due to point charges when
(i) $q<0$ and (ii) $q>0 . \quad$
View full question & answer→(i) $q<0$ and (ii) $q>0 . \quad$
Question 41 Mark
Can a metal sphere of radius 1 em be given a charge of IC?
Answer
View full question & answer→No, electric field at the bottom of the sphere
$E =\frac{1}{4 \pi \epsilon_0} \times \frac{q}{ R ^2}$
$=9 \times 10^9 \times \frac{1}{(0.01)^2}$
$=9 \times 10^{13} V / m$
If the electric field intensity in air is $3 \times 10^6 V / m$, air will be ionized and the charge of the sphere will be destroyed in air Therefore the sphere will not be able to hold a charge of 1 C.
$E =\frac{1}{4 \pi \epsilon_0} \times \frac{q}{ R ^2}$
$=9 \times 10^9 \times \frac{1}{(0.01)^2}$
$=9 \times 10^{13} V / m$
If the electric field intensity in air is $3 \times 10^6 V / m$, air will be ionized and the charge of the sphere will be destroyed in air Therefore the sphere will not be able to hold a charge of 1 C.
Question 51 Mark
If an electric dipole is placed at an angle of $30^{\circ}$ to an electric field of intensity $2 \times 10^5 N / C$, a torque of 4 Nm is applied on it. If the length of the dipole is 2 cm , then what will be charge?
Answer
View full question & answer→ $\tau= pE \sin \theta$
$4=p \times 2 \times 10^5 \sin 30^{\circ}$
$p=4 \times 10^{-5} C - m$
$q \times 2 \times 10^{-2}=4 \times 10^{-5}$
$q=2 \times 10^{-3} C$
$\because p=q \times 2 l$
$4=p \times 2 \times 10^5 \sin 30^{\circ}$
$p=4 \times 10^{-5} C - m$
$q \times 2 \times 10^{-2}=4 \times 10^{-5}$
$q=2 \times 10^{-3} C$
$\because p=q \times 2 l$
Question 61 Mark
A sample of $\ce{HCl}$ gas is placed in an electric field of $3 \times 10^4 N / C$. The dipole moment of each molecule of $\ce{HCl}$ is $6 \times 10^{-30} C \times m$. What will be the maximum torque acting on the molecule?
Answer
View full question & answer→Torque on dipole $\tau= P \times E$
$ \tau=P E \sin \theta $
For maximum value, $\theta=90^{\circ}$
$ \tau_{\max }=6 \times 10^{-30} \times 3 \times 10^4 \sin 90^{\circ}$
$\tau_{\max }=18 \times 10^{-26} Nm $
$ \tau=P E \sin \theta $
For maximum value, $\theta=90^{\circ}$
$ \tau_{\max }=6 \times 10^{-30} \times 3 \times 10^4 \sin 90^{\circ}$
$\tau_{\max }=18 \times 10^{-26} Nm $
Question 71 Mark
Write two differences between charge and mass.
Answer
View full question & answer→(1) Charge can be zero, positive or negative but mass is always positive.
(ii) Electric charge does not depend on the velocity of motion of the particle whereas mass also depends on the velocity of motion of the particle.
(ii) Electric charge does not depend on the velocity of motion of the particle whereas mass also depends on the velocity of motion of the particle.
Question 81 Mark
Two protons and two electrons are hanging freely at equal distances. Compare the repulsion force between proton and electron.
Answer
View full question & answer→The repulsion forces between two protons and two electrons will be equal because the magnitude of charge on the proton and electron is equal and the distances between them are also equal.
Question 91 Mark
If we walk barefoot on a nylon mat for some time and touch the metal handle of a door, we get an electric shock. Why does this happen?
Answer
View full question & answer→While walking, our feet rub on the mat, due to which our body gets charged. When we touch the handle, electric charge from our body flows into it, due to which we
get a shock.
Question 101 Mark
Two point charges are placed at a distance d from each other. Due to these, the electric field intensity at any point is zero. But the point is not in the middle of these charges although on the line connecting them, write two necessary conditions for this to happen.
Answer
View full question & answer→(i) Both the charges should be of opposite nature.(ii) The magnitude of the charge located near that point should be less and the magnitude of the other charge should be greater.
Question 111 Mark
An electron and a proton are placed in a uniform electric field. Which will have greater acceleration and why?
Answer
View full question & answer→Electron, because the mass of electron is less.
Question 121 Mark
What will be the change in the force acting on a point charge placed on its axis if its distance is doubled?
Answer
View full question & answer→Force $F =q E$
$=q \times \frac{1}{4 \pi \epsilon_0} \times \frac{2 p}{r^3}$
Hence, the force will remain $\frac{1}{8}$ th of the previous value.
$=q \times \frac{1}{4 \pi \epsilon_0} \times \frac{2 p}{r^3}$
Hence, the force will remain $\frac{1}{8}$ th of the previous value.
Question 131 Mark
There is a potential of 10V on the surface of a charged spherical shell of radius 10 cm. Write the value of potential at a distance of 5 cm from its centre.
Answer
View full question & answer→The potential is the same at all points within the spherical surface and its value is the same as that on the surface of a spherical charge, that is, the value of the potential at a distance of 5 cm from its centre will be 10 Volt.
Question 141 Mark
An electric dipole of charge $\pm 1 \mu C$ is located inside a spherical Gaussian surface of radius 1 cm . Write the value of electric flux emerging from the Gaussian surface.
Answer
View full question & answer→Electric flux emerging out of Gaussian surface $\phi=\frac{q}{\epsilon_0} $Here, $\epsilon_0$ is the electrical conductivity of air or vacuumIn the given question the algebraic sum of charges is zero.
Therefore the value of electric flux emmiting from the Gaussian surface will be zero.
Therefore the value of electric flux emmiting from the Gaussian surface will be zero.
Question 151 Mark
Write any two properties of electric field lines.
Answer
View full question & answer→(i) The tangent drawn at any point to the line of electric force expresses the direction of the resultant electric field at that point.
(ii) Electric lines of force run from positive charge and end at netative charge.
Note: Students can also write other properties of electric field lines.
(ii) Electric lines of force run from positive charge and end at netative charge.
Note: Students can also write other properties of electric field lines.
Question 161 Mark
Can a metal sphere of radius one meter be given a charge of 1 coulomb?
Answer
View full question & answer→No, because
$\begin{aligned}
E= \frac{k q}{R^2}=\frac{9 \times 10^9 \times 1}{1^2} \\
E=9 \times 10^9 \frac{\text { Volt }}{\text { meter }}\end{aligned}$
will occur on the surface of the sphere.
If the electric field in the air exceods $3 \times 10^6$ Volt/ meter, the air will get ionized due to which the charge of the sphere will get destroyed in the air.
$\begin{aligned}
E= \frac{k q}{R^2}=\frac{9 \times 10^9 \times 1}{1^2} \\
E=9 \times 10^9 \frac{\text { Volt }}{\text { meter }}\end{aligned}$
will occur on the surface of the sphere.
If the electric field in the air exceods $3 \times 10^6$ Volt/ meter, the air will get ionized due to which the charge of the sphere will get destroyed in the air.
Question 171 Mark
Write the formula to find the value of electric field due to a uniformly charged infinite plane sheet.
Answer
View full question & answer→$\overrightarrow{ E }=\frac{\sigma}{2 \epsilon_0} \hat{n}$
Here $\hat{n}$ is the unit vector moving away from it perpendicular to the plane.
Here $\hat{n}$ is the unit vector moving away from it perpendicular to the plane.
Question 181 Mark
How does the electric field due to point charge and line charge change with distance?
Answer
View full question & answer→Due to point charge
$E =\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{r^2}\right)$
$E \propto \frac{1}{r^2}$
And due to linear charge $E =\frac{\lambda}{2 \pi \epsilon_0 r} \Rightarrow E \propto \frac{1}{r}$
$E =\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{r^2}\right)$
$E \propto \frac{1}{r^2}$
And due to linear charge $E =\frac{\lambda}{2 \pi \epsilon_0 r} \Rightarrow E \propto \frac{1}{r}$
Question 191 Mark
The flux inside a balloon of radius 7.0 cm is 2.0 - $10^3 NmC ^{-1}$. If the radius of the balloon is doubled, what will be the value of flux?
Answer
View full question & answer→This value will remain unchanged because the charge enclosed by the Gaussian surface remains the same.
Question 201 Mark
A Gaussian surface has change $-q+3 q$ and $-2 q$. Calcualte the resultant electric flux through the surface.
Answer
View full question & answer→Total charge enclosed by the Gaussian surface
$ \begin{aligned} & =-q+3 q-2 q=0 \\
\phi & =\frac{q}{\epsilon_0}=0 \quad
\because q=0 \end{aligned} $
$ \begin{aligned} & =-q+3 q-2 q=0 \\
\phi & =\frac{q}{\epsilon_0}=0 \quad
\because q=0 \end{aligned} $
Question 211 Mark
Write the value of electric flux due to a point charge located outside the surface of a sphere.
Answer
View full question & answer→From Gauss theorem $\phi=\frac{q}{\epsilon_0}$ $\because q=0$, therefore $\phi=0$
Question 221 Mark
The value of electric flux through a closed surface is zero. What does this mean?
Answer
View full question & answer→It means that the value of electric flux towards inside is equal to the electric flux towards outside.
Question 231 Mark
Can a solid conducting sphere of the same radius be charged more than a hollow sphere? Give the reason also.
Answer
View full question & answer→No, because the entire charge of a charged conductor is always on the outer surface of the conductor.
Question 241 Mark
Does the value of electric flux depend on the shape of the Gaussian surface?
Answer
View full question & answer→No
Question 251 Mark
What is the importance of Gaussian surface?
Answer
View full question & answer→The principle of Gaussian surface helps in finding the intensity of electric field due to symmetric charge distribution.
Question 261 Mark
What is a Gaussian surface?
Answer
View full question & answer→It is a region around a charge in which the electric field intensity is equal at all points and the electric flux is always perpendicular to the surface.
Question 271 Mark
What is Gauss' law?
Answer
View full question & answer→The total value of electric flux passing through a closed surface in free space is $\frac{1}{\epsilon_0}$ times the charge enclosed by the surface.
Electric flux passing through a closed surface S
$ \phi=\frac{q}{\epsilon_0} $
Here $q$, is the total charge enclosed bv the surface S .
Electric flux passing through a closed surface S
$ \phi=\frac{q}{\epsilon_0} $
Here $q$, is the total charge enclosed bv the surface S .
Question 281 Mark
What is the magnitude of the torque and also give its direction.
Answer
View full question & answer→Magnitude of torque $=q E \cdot 2 a \sin \theta$
$=2 q a E \sin \theta$
The direction of $\vec{p}$ normal to the plane of $\overrightarrow{ E }$ is given by the right hand screw rule. But $p=2 a q$
$\therefore \quad$ Magnitude of torque $=p E \sin \theta$
$=2 q a E \sin \theta$
The direction of $\vec{p}$ normal to the plane of $\overrightarrow{ E }$ is given by the right hand screw rule. But $p=2 a q$
$\therefore \quad$ Magnitude of torque $=p E \sin \theta$
Question 291 Mark
When an electric dipole is placed in a non- uniform electric field, does it experience a force?
Answer
View full question & answer→Yes, it experiences a force in a non-uniform electric field.
Question 301 Mark
What is the relationship between the electric field intensities due to a small electric dipole on its axis and abscissa?
Answer
View full question & answer→$E _{\text {axial }}=2 E _{\text {uniaxial }}$
Question 311 Mark
What is the resultant force on an electric dipole placed in a uniform electric field?
Answer
View full question & answer→Zero
Question 321 Mark
What is the direction of the electric dipole moment vector quantity?
Answer
View full question & answer→From negative to positive charge.
Question 331 Mark
A dipole of dipole moment $\vec{P}$ is placed in a uniform electric field $\vec{E}$. Write the value of the angle between $\vec{P}$ and $\vec{E}$ for which the torque experienced by the dipole is minimum.
Answer
View full question & answer→$\theta=0^{\circ} \text { (because } t=P E \sin \theta \text { for } \tau_{\min } \left. \sin \theta=0, \theta=0^{\circ}\right)$
Question 341 Mark
In which configuration will an electric dipole be in (i) permanent and (ii) temporary equilibrium when placed under a uniform electric field?
Answer
View full question & answer→(i) When the electric dipole moment $\overrightarrow{ P }$ and the electric field intensity $\vec{E}$ have the same direction.
(ii) When $\overrightarrow{ P }$ and $\overrightarrow{ E }$ are in opposite directions.
(ii) When $\overrightarrow{ P }$ and $\overrightarrow{ E }$ are in opposite directions.
Question 351 Mark
Define electric dipole. Write its unit.
Answer
View full question & answer→A pair of very close point charges, equal in magnitude and opposite in nature, is called an electric dipole. Its unit is Coulomb meter.
Question 361 Mark
What is the nature of electric lines of force in a uniform electric field?
Answer
View full question & answer→In a uniform electric field, the lines of force are straight lines, parallel and at equal distances. As shown in the Fig
Also see question 30. i.e. the non-uniform electric field.
Also see question 30. i.e. the non-uniform electric field.
Question 371 Mark
Define electric flux.
Answer
View full question & answer→The electric flux $\overrightarrow{\Delta S}$ passing through an area element $\Delta \phi$ is detined as follows :
$ \Delta \phi=\vec{E} \cdot \overrightarrow{\Delta S}=E \Delta S \cos \theta $
Here, $\theta$ is the angle between the area element $\overrightarrow{\Delta S}$ and $\overrightarrow{ E }$
$ \Delta \phi=\vec{E} \cdot \overrightarrow{\Delta S}=E \Delta S \cos \theta $
Here, $\theta$ is the angle between the area element $\overrightarrow{\Delta S}$ and $\overrightarrow{ E }$
Question 381 Mark
What type of electric field is shown by the lines in the figure?
Answer
View full question & answer→Both the magnitude and direction of the electric field are variable. i.e. the electric field is non-uniform.
Question 391 Mark
Why can't two lines of electric force intersect each other?
Answer
View full question & answer→Lines of force do not intersect each other, because if they intersect then two tangent lines at the intersection point will express two resultant fields, which is impossible.
Question 401 Mark
Definition of electric field intensity at a point.
Answer
View full question & answer→The electric force acting on a unit test positive charge at any point in the electric field is called the intensity of the electric field at that point.
$E =\lim _{q_0 \rightarrow 0}\left(\frac{F}{q_0}\right)$
$E =\lim _{q_0 \rightarrow 0}\left(\frac{F}{q_0}\right)$
Question 411 Mark
Give the physical quantity whose unit is Newton/Coulomb.
Answer
View full question & answer→Electric field intensity
Question 421 Mark
Is the electric field intensity a scalar or a vector quantity?
Answer
View full question & answer→Electric field intensity is a vector quantity
Question 431 Mark
What is the force acting on a charge Q placed in an electric field E?
Answer
View full question & answer→$\vec{F}=\overrightarrow{Q E}$
Question 441 Mark
What is the value of the electric field produced by charge $Q$ at any point $\vec{r}$ ?
Answer
View full question & answer→$\overrightarrow{ E }=\frac{1}{4 \pi \epsilon_0} \frac{ Q }{r^2} \vec{r}$
Question 451 Mark
What is the value of dielectric constant of metals?
Answer
View full question & answer→Infinite
Question 461 Mark
Two equal point charges are placed at a dis- tance of 1 meter from each other, experience a force of 8N. If these charges are kept in water at the same distance, then how much force will be felt between them?
Answer
View full question & answer→We know$F_{m}=\frac{F_{a}}{m}=\frac{8 N}{80}=0.1 N$
Question 471 Mark
Define the dielectric constant of a medium.
Answer
View full question & answer→The ratio of absolute permittivity of a medium and permittivity of air or vacuum is called dielectric constant (relative permittivity) of the medium.
Therefore, $\epsilon_{ r }=\frac{\epsilon}{\epsilon_0}$ or $\epsilon=\epsilon_{ o } \epsilon_{ T }$
Therefore, $\epsilon_{ r }=\frac{\epsilon}{\epsilon_0}$ or $\epsilon=\epsilon_{ o } \epsilon_{ T }$
Question 481 Mark
If two electric charges $q_1$ and $q_2$ are such that (i) $q_1 q_2>0$, (ii) $q_1 q_2<0$, then what will be the nature of the electric force between them?
Answer
View full question & answer→(i) Repulsive (since in this case both the chargers will be of similar nature).
(ii) Attractive (since in this case both the charges will be of opposite nature).
Question 491 Mark
How will the force between two point charges change, if the dielectric constant of the medium in which they are kept increases?
Answer
View full question & answer→If the dielectric constant of any one medium increases. So the force between the point charges will decrease.
Question 501 Mark
Write Coulomb's law.
Answer
View full question & answer→The force of attraction or repulsion (F) between two point charges is proportional to the product of the magnitude of the charges $q_1$ and $q_2$ and inversely proportional to the square of the distance between the charges $\left(1 / r^2\right)$.
That is $F \propto \frac{q_1 q_2}{r^2}$
or $F = K \frac{q_1 q_2}{r^2}$
or $F =\frac{1}{4 \pi \epsilon_o} \frac{q_1 q_2}{r^2}$
Here, $K \left(=\frac{1}{4 \pi \epsilon_{ o }}\right)=9 \times 10^9 Nm ^2 / C ^2$.
That is $F \propto \frac{q_1 q_2}{r^2}$
or $F = K \frac{q_1 q_2}{r^2}$
or $F =\frac{1}{4 \pi \epsilon_o} \frac{q_1 q_2}{r^2}$
Here, $K \left(=\frac{1}{4 \pi \epsilon_{ o }}\right)=9 \times 10^9 Nm ^2 / C ^2$.