Answer

Repulsive force of magnitude $6 \times 10^{-3} \mathrm{~N}$

Charge on the first sphere, $q_{1}=2 \times 10^{-7} \mathrm{C}$

Charge on the second sphere, $q_{2}=3 \times 10^{-7} \mathrm{C}$

Distance between the spheres, $r=30 \mathrm{~cm}=0.3 \mathrm{~m}$

Electrostatic force between the spheres is given by the relation,

$$ F=\frac{q_{1} q_{2}}{4 \pi \in_{0} r^{2}} $$

Where, $\epsilon_{0}=$ Permittivity of free space

$$ \begin{aligned} & \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2} \\ & F=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} \mathrm{~N} \end{aligned} $$

Hence, force between the two small charged spheres is $6 \times 10^{-3} \mathrm{~N}$. The charges are of same nature. Hence, force between them will be repulsive.

Answer

Electrostatic force on the first sphere, $F=0.2 \mathrm{~N}$

Charge on this sphere, $q_{1}=0.4 \mu \mathrm{C}=0.4 \times 10^{-6} \mathrm{C}$

Charge on the second sphere, $q_{2}=-0.8 \mu \mathrm{C}=-0.8 \times 10^{-6} \mathrm{C}$

Electrostatic force between the spheres is given by the relation,

$$ F=\frac{q_{1} q_{2}}{4 \pi \in_{0} r^{2}} $$

Where, $\epsilon_{0}=$ Permittivity of free space

$$ \begin{aligned} & \text { And, } \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2} \\ & \begin{aligned} r^{2} & =\frac{q_{1} q_{2}}{4 \pi \in_{0} F} \\ & =\frac{0.4 \times 10^{-6} \times 8 \times 10^{-6} \times 9 \times 10^{9}}{0.2} \\ & =144 \times 10^{-4} \\ r & =\sqrt{144 \times 10^{-4}}=0.12 \mathrm{~m} \end{aligned} \end{aligned} $$

The distance between the two spheres is $0.12 \mathrm{~m}$.

Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is $0.2 \mathrm{~N}$.

Answer

The given ratio is $\frac{k e^{2}}{\mathrm{G} m_{\mathrm{e}} m_{\mathrm{p}}}$.

Where,

$\mathrm{G}=$ Gravitational constant

Its unit is $\mathrm{N} \mathrm{m}^{2} \mathrm{~kg}^{-2}$.

$m_{\mathrm{e}}$ and $m_{\mathrm{p}}=$ Masses of electron and proton.

Their unit is $\mathrm{kg}$.

$e=$ Electric charge.

Its unit is C.

$k=\mathrm{A}$ constant

$=\frac{1}{4 \pi \epsilon_{0}}$

$\epsilon_{0}=$ Permittivity of free space

Its unit is $\mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-2}$.

Therefore, unit of the given ratio $\frac{k e^{2}}{\mathrm{G} m_{\mathrm{e}} m_{\mathrm{p}}}=\frac{\left[\mathrm{Nm}^{2} \mathrm{C}^{-2}\right]\left[\mathrm{C}^{-2}\right]}{\left[\mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right][\mathrm{kg}][\mathrm{kg}]}$ $=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$

Hence, the given ratio is dimensionless.

$e=1.6 \times 10^{-19} \mathrm{C}$

$\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}$

$m_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}$

$m_{\mathrm{p}}=1.66 \times 10^{-27} \mathrm{~kg}$

Hence, the numerical value of the given ratio is

$\frac{k e^{2}}{\mathrm{G} m_{e} m_{p}}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-3} \times 1.67 \times 10^{-22}} \approx 2.3 \times 10^{39}$

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Answer

Electric charge of a body is quantized. This means that only integral $(1,2, \ldots, n)$ number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Answer

Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Answer

The given figure shows a square of side $10 \mathrm{~cm}$ with four charges placed at its corners. $\mathrm{O}$ is the centre of the square.

Where,

(Sides) $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}=10 \mathrm{~cm}$

(Diagonals) $\mathrm{AC}=\mathrm{BD}=10 \sqrt{2} \mathrm{~cm}$

$\mathrm{AO}=\mathrm{OC}=\mathrm{DO}=\mathrm{OB}=5 \sqrt{2} \mathrm{~cm}$

A charge of amount $1 \mu \mathrm{C}$ is placed at point $\mathrm{O}$.

Force of repulsion between charges placed at corner $\mathrm{A}$ and centre $\mathrm{O}$ is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner $\mathrm{C}$ and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner $\mathrm{B}$ and centre $\mathrm{O}$ is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner $\mathrm{D}$ and centre $\mathrm{O}$. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on $1 \mu \mathrm{C}$ charge at centre $\mathrm{O}$ is zero.

Answer

An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Answer

The situation is represented in the given figure. $\mathrm{O}$ is the mid-point of line $\mathrm{AB}$.

Distance between the two charges, $\mathrm{AB}=20 \mathrm{~cm}$ $\therefore \mathrm{AO}=\mathrm{OB}=10 \mathrm{~cm}$

Net electric field at point $\mathrm{O}=E$

Electric field at point $\mathrm{O}$ caused by $+3 \mu \mathrm{C}$ charge,

$E_{1}=\frac{3 \times 10^{-6}}{4 \pi \epsilon_{0}(\mathrm{AO})^{2}}=\frac{3 \times 10^{-6}}{4 \pi \epsilon_{0}\left(10 \times 10^{-2}\right)^{2}} \mathrm{~N} / \mathrm{C} \quad$ along $\mathrm{OB}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$

Magnitude of electric field at point $\mathrm{O}$ caused by $-3 \mu \mathrm{C}$ charge,

$E_{2}=\left|\frac{-3 \times 10^{-6}}{4 \pi \epsilon_{0}(\mathrm{OB})^{2}}\right|=\frac{3 \times 10^{-6}}{4 \pi \epsilon_{0}\left(10 \times 10^{-2}\right)^{2}} \mathrm{~N} / \mathrm{C} \quad$ along $\mathrm{OB}$

$\therefore E=E_{1}+E_{2}$

$=2 \times\left[\left(9 \times 10^{9}\right) \times \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}}\right] \quad\left[\right.$ Since the values of $E_{1}$ and $E_{2}$ are same, the value is

multiplied with 2]

$=5.4 \times 10^{6} \mathrm{~N} / \mathrm{C}$ along $\mathrm{OB}$

Therefore, the electric field at mid-point $\mathrm{O}$ is $5.4 \times 10^{6} \mathrm{~N} \mathrm{C}^{-1}$ along $\mathrm{OB}$.

A test charge of amount $1.5 \times 10^{-9} \mathrm{C}$ is placed at mid-point $\mathrm{O}$.

$q=1.5 \times 10^{-9} \mathrm{C}$

Force experienced by the test charge $=F$

$\therefore F=q E$

$=1.5 \times 10^{-9} \times 5.4 \times 10^{6}$

$=8.1 \times 10^{-3} \mathrm{~N}$

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point $B$ but attracted towards point $A$.

Therefore, the force experienced by the test charge is $8.1 \times 10^{-3} \mathrm{~N}$ along $\mathrm{OA}$.

Answer

Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, $q_{\mathrm{A}}=2.5 \times 10^{-7} \mathrm{C}$

At B, amount of charge, $q_{\mathrm{B}}=-2.5 \times 10^{-7} \mathrm{C}$

Total charge of the system,

$q=q_{\mathrm{A}}+q_{\mathrm{B}}$

$=2.5 \times 10^{7} \mathrm{C}-2.5 \times 10^{-7} \mathrm{C}$

$=0$

Distance between two charges at points A and B,

$d=15+15=30 \mathrm{~cm}=0.3 \mathrm{~m}$

Electric dipole moment of the system is given by, $p=q_{\mathrm{A}} \times d=q_{\mathrm{B}} \times d$

$=2.5 \times 10^{-7} \times 0.3$

$=7.5 \times 10^{-8} \mathrm{C} \mathrm{m}$ along positive $z$-axis

Therefore, the electric dipole moment of the system is $7.5 \times 10^{-8} \mathrm{C} \mathrm{m}$ along positive $z$-axis.

Answer

Electric dipole moment, $p=4 \times 10^{-9} \mathrm{C} \mathrm{m}$

Angle made by $p$ with a uniform electric field, $\theta=30^{\circ}$

Electric field, $E=5 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}$

Torque acting on the dipole is given by the relation,

$\tau=p E \sin \theta$

$=4 \times 10^{-9} \times 5 \times 10^{4} \times \sin 30$

$=20 \times 10^{-5} \times \frac{1}{2}$

$=10^{-4} \mathrm{~N} \mathrm{~m}$

Therefore, the magnitude of the torque acting on the dipole is $10^{-4} \mathrm{~N} \mathrm{~m}$.

Answer

When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.

Amount of charge on the polythene piece, $q=-3 \times 10^{-7} \mathrm{C}$

Amount of charge on an electron, $e=-1.6 \times 10^{-19} \mathrm{C}$

Number of electrons transferred from wool to polythene $=n$

$n$ can be calculated using the relation,

$q=n e$

$n=\frac{q}{e}$

$=\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}}$

$=1.87 \times 10^{12}$

Therefore, the number of electrons transferred from wool to polythene is $1.87 \times 10^{12}$.

Yes.

There is a transfer of mass taking place. This is because an electron has mass,

$m_{e}=9.1 \times 10^{-3} \mathrm{~kg}$

Total mass transferred to polythene from wool,

$m=m_{e} \times n$

$=9.1 \times 10^{-31} \times 1.85 \times 10^{12}$

$=1.706 \times 10^{-18} \mathrm{~kg}$

Hence, a negligible amount of mass is transferred from wool to polythene.

Answer

Charge on sphere A, $q_{\mathrm{A}}=$ Charge on sphere B, $q_{\mathrm{B}}=6.5 \times 10^{-7} \mathrm{C}$

Distance between the spheres, $r=50 \mathrm{~cm}=0.5 \mathrm{~m}$

Force of repulsion between the two spheres,

$$ F=\frac{q_{\mathrm{A}} q_{\mathrm{B}}}{4 \pi \in_{0} r^{2}} $$

Where,

$\epsilon_{0}=$ Free space permittivity

$$ \begin{aligned} & \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2} \\ & \quad F=\frac{9 \times 10^{9} \times\left(6.5 \times 10^{-7}\right)^{2}}{(0.5)^{2}} \\ & \therefore \quad \\ & =1.52 \times 10^{-2} \mathrm{~N} \end{aligned} $$

Therefore, the force between the two spheres is $1.52 \times 10^{-2} \mathrm{~N}$.

After doubling the charge, charge on sphere A, $q_{\mathrm{A}}=$ Charge on sphere B, $q_{\mathrm{B}}=2 \times 6.5 \times$ $10^{-7} \mathrm{C}=1.3 \times 10^{-6} \mathrm{C}$

The distance between the spheres is halved.

$$ \therefore \quad r=\frac{0.5}{2}=0.25 \mathrm{~m} $$

Force of repulsion between the two spheres,

$$ \begin{aligned} & F=\frac{q_{\mathrm{A}} q_{\mathrm{B}}}{4 \pi \epsilon_{0} r^{2}} \\ & =\frac{9 \times 10^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^{2}} \\ & =16 \times 1.52 \times 10^{-2} \\ & =0.243 \mathrm{~N} \end{aligned} $$

Therefore, the force between the two spheres is $0.243 \mathrm{~N}$.

Answer

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Answer

Electric field intensity, $\vec{E}=3 \times 10^{3} \hat{\imath} \mathrm{N} / \mathrm{C}$

Magnitude of electric field intensity, $|\vec{E}|_{=3 \times 10^{3} \mathrm{~N} / \mathrm{C}}$

Side of the square, $s=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Area of the square, $A=\mathrm{s}^{2}=0.01 \mathrm{~m}^{2}$

The plane of the square is parallel to the $y-z$ plane. Hence, angle between the unit vector normal to the plane and electric field, $\theta=0^{\circ}$

Flux $(\Phi)$ through the plane is given by the relation,

$\Phi=|\vec{E}| A \cos \theta$

$=3 \times 10^{3} \times 0.01 \times \cos 0^{\circ}$

$=30 \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

Plane makes an angle of $60^{\circ}$ with the $x$-axis. Hence, $\theta=60^{\circ}$

Flux, $\Phi=|\vec{E}| A \cos \theta$

$=3 \times 10^{3} \times 0.01 \times \cos 60^{\circ}$ $=30 \times \frac{1}{2}=15 \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

Answer

All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Answer

(a) Net outward flux through the surface of the box, $\Phi=8.0 \times 10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

For a body containing net charge $q$, flux is given by the relation,

$\phi=\frac{q}{\epsilon_{0}}$

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$ $q=\epsilon_{0} \Phi$

$=8.854 \times 10^{-12} \times 8.0 \times 10^{3}$

$=7.08 \times 10^{-8}$

$=0.07 \mu \mathrm{C}$

Therefore, the net charge inside the box is $0.07 \mu \mathrm{C}$.

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Answer

The square can be considered as one face of a cube of edge $10 \mathrm{~cm}$ with a centre where charge $q$ is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

$$ \phi_{\text {Total }}=\frac{q}{\epsilon_{0}} $$

Hence, electric flux through one face of the cube i.e., through the square, $\phi=\frac{\phi_{\text {Total }}}{6}$ $=\frac{1}{6} \frac{q}{\epsilon_{0}}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$q=10 \mu \mathrm{C}=10 \times 10^{-6} \mathrm{C}$

$\therefore \phi=\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}$

$=1.88 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}$

Therefore, electric flux through the square is $1.88 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}$.

Answer

Net electric flux ( $\left.\Phi_{\mathrm{Net}}\right)$ through the cubic surface is given by,

$$ \phi_{\mathrm{Net}}=\frac{q}{\epsilon_{0}} $$

Where,

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$q=$ Net charge contained inside the cube $=2.0 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}$ $\therefore \phi_{\text {Net }}=\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}$

$=2.26 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}$

The net electric flux through the surface is $2.26 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}$.

Answer

Electric flux, $\Phi=-1.0 \times 10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

Radius of the Gaussian surface,

$r=10.0 \mathrm{~cm}$

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., $-10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$.

Electric flux is given by the relation,

$\phi=\frac{q}{\epsilon_{0}}$

Where,

$q=$ Net charge enclosed by the spherical surface

$\epsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$\therefore q=\phi \in_{0}$

$=-1.0 \times 10^{3} \times 8.854 \times 10^{-12}$ $=-8.854 \times 10^{-9} \mathrm{C}$

$=-8.854 \mathrm{nC}$

Therefore, the value of the point charge is $-8.854 \mathrm{nC}$.

Answer

Electric field intensity $(E)$ at a distance $(d)$ from the centre of a sphere containing net charge $q$ is given by the relation,

$$ E=\frac{q}{4 \pi \in_{0} d^{2}} $$

Where,

$q=$ Net charge $=1.5 \times 10^{3} \mathrm{~N} / \mathrm{C}$

$d=$ Distance from the centre $=20 \mathrm{~cm}=0.2 \mathrm{~m}$

$\epsilon_{0}=$ Permittivity of free space

And, $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}$

$\therefore q=E\left(4 \pi \epsilon_{0}\right) d^{2}$

$=\frac{1.5 \times 10^{3} \times(0.2)^{2}}{9 \times 10^{9}}$

$=6.67 \times 10^{9} \mathrm{C}$

$=6.67 \mathrm{nC}$

Therefore, the net charge on the sphere is $6.67 \mathrm{nC}$.

Answer

Diameter of the sphere, $d=2.4 \mathrm{~m}$

Radius of the sphere, $r=1.2 \mathrm{~m}$

Surface charge density, $\sigma=80.0 \mu \mathrm{C} / \mathrm{m}^{2}=80 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}$

Total charge on the surface of the sphere,

$Q=$ Charge density $\times$ Surface area

$=\sigma \times 4 \pi r^{2}$

$=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^{2}$

$=1.447 \times 10^{-3} \mathrm{C}$

Therefore, the charge on the sphere is $1.447 \times 10^{-3} \mathrm{C}$.

Total electric flux ( $\phi_{\text {Total }}$ ) leaving out the surface of a sphere containing net charge $Q$ is given by the relation,

$$ \phi_{\text {TOtal }}=\frac{Q}{\epsilon_{0}} $$

Where,

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$$ \begin{aligned} & Q=1.447 \times 10^{-3} \mathrm{C} \\ & \phi_{\text {Total }}=\frac{1.44 \times 10^{-3}}{8.854 \times 10^{-12}} \\ & =1.63 \times 10^{8} \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2} \end{aligned} $$

Therefore, the total electric flux leaving the surface of the sphere is $1.63 \times 10^{8} \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}$.

Answer

Electric field produced by the infinite line charges at a distance $d$ having linear charge density $\lambda$ is given by the relation,

$$ \begin{aligned} & E=\frac{\lambda}{2 \pi \epsilon_{0} d} \\ & \lambda=2 \pi \epsilon_{0} d E \end{aligned} $$

Where,

$d=2 \mathrm{~cm}=0.02 \mathrm{~m}$

$E=9 \times 10^{4} \mathrm{~N} / \mathrm{C}$

$\epsilon_{0}=$ Permittivity of free space

$$ \begin{aligned} & \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2} \\ & \lambda=\frac{0.02 \times 9 \times 10^{4}}{2 \times 9 \times 10^{9}} \\ & =10 \mu \mathrm{C} / \mathrm{m} \end{aligned} $$

Therefore, the linear charge density is $10 \mu \mathrm{C} / \mathrm{m}$.

Answer

Let $A$ and $B$ be two large, thin metal plates held parallel and close to each other as shown in surface density of charges on $A \sigma=+17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2$. And surface density of charges on $B,-\sigma=-17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2$ At any point $\mathrm{K}$ in the outer region of first plate $\mathrm{A}$, we have $\left|\overrightarrow{E_A}\right|=\left|\overrightarrow{E_B}\right|=\frac{\sigma}{2 \epsilon_0}$ but their directions are mutually opposite,

Hence $\quad$ Overseo $\left(E_k\right)=\overrightarrow{E_A}+\overrightarrow{E_B}=\overrightarrow{0}$ (b) Let $\mathrm{A}$ and $\mathrm{B}$ be two large, thin metal plates held parallel and close to each other as shown in surface density of charges on $A \sigma=+17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2$. And surface density of charges on

$ B,-\sigma=-17.0 \times 10^{-22} C / m^2 $

Again at a point $M$ in the outer region of the second plate, net electric field is zero (c) Let $\mathrm{A}$ and $\mathrm{B}$ be two large, thin metal plates held parallel and close to each other as shown in surface density of charges on $A \sigma=+17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2$. And surface density of charges on

$ B,-\sigma=-17.0 \times 10^{-22} C / m^2 $

At a point $\mathrm{N}$ between the plates $\mathrm{A}$ and $\mathrm{B}$, and $\overrightarrow{E_B}$ are equal in magnitude and both are directed in same direction and are, therefore, added up, hence net electric field is given by,

$$ \begin{aligned} & \vec{E}=\overrightarrow{E_A}+\overrightarrow{E_B}=\frac{\sigma}{2 \epsilon_0}+\frac{\sigma}{2 \epsilon_0}=\frac{\sigma}{\epsilon_0} \\ & =\frac{1.70 \times 10^{-22}}{8.85 \times 10^{-12}}=1.9 \times 10^{-10} N C^{-1} \end{aligned} $$