Class 12 Physics – Chapter 1: Electric Charges and Fields

NCERT Exercise Solutions 1.1–1.23

The following solutions follow the standard NCERT Class 12 Physics Chapter 1 approach. (NCERT)

Note: In the text you pasted, several units such as μC appear to have been lost during copying. I have used the standard NCERT values/units where necessary.

1.1 Force between two charged spheres

Given:

$$q_1=2\times10^{-7}C$$ $$q_2=3\times10^{-7}C$$ $$r=30\,cm=0.30\,m$$

Using Coulomb's law:

$$F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$

where

$$k=\frac{1}{4\pi\epsilon_0}=9\times10^9\,Nm^2/C^2$$

Therefore,

$$F=\frac{9\times10^9(2\times10^{-7})(3\times10^{-7})}{(0.30)^2}$$ $$F=\frac{54\times10^{-5}}{0.09}$$ $$\boxed{F=6\times10^{-3}N}$$

Since both charges are positive, the force is repulsive.

Answer:

$$\boxed{6\times10^{-3}N}$$

1.2 Electrostatic force between two charged spheres

Given:

$$q_1=0.4\,\mu C=0.4\times10^{-6}C$$ $$q_2=-0.8\,\mu C$$ $$F=0.2N$$

(a) Distance between the spheres

Using

$$F=\frac{k|q_1q_2|}{r^2}$$

Therefore,

$$r=\sqrt{\frac{k|q_1q_2|}{F}}$$ $$r=\sqrt{\frac{9\times10^9(0.4\times10^{-6})(0.8\times10^{-6})}{0.2}}$$ $$r=\sqrt{0.0144}$$ $$\boxed{r=0.12m}$$

or

$$\boxed{r=12cm}$$

(b) Force on the second sphere

According to Newton's third law, the force on the second sphere is equal in magnitude and opposite in direction.

$$\boxed{F=0.2N}$$

Because the charges are opposite, the force is attractive.

1.3 Ratio $\frac{ke^2}{Gm_em_p}$

We have:

$$\frac{ke^2}{Gm_em_p}$$

where

• $k=9\times10^9\,Nm^2/C^2$

• $e=1.602\times10^{-19}C$

• $G=6.674\times10^{-11}Nm^2/kg^2$

• $m_e=9.109\times10^{-31}kg$

• $m_p=1.673\times10^{-27}kg$

Substitution gives

$$\frac{ke^2}{Gm_em_p}\approx2.27\times10^{39}$$ $$\boxed{\frac{ke^2}{Gm_em_p}\approx2.3\times10^{39}}$$

Why is it dimensionless?

Electric force:

$$F_e=\frac{ke^2}{r^2}$$

Gravitational force:

$$F_g=\frac{Gm_em_p}{r^2}$$

Thus,

$$\frac{F_e}{F_g} = \frac{ke^2}{Gm_em_p}$$

Both numerator and denominator represent force, so their ratio has no dimensions.

Significance

The ratio means that the electrostatic force between an electron and proton is about $2.3\times10^{39}$ times stronger than their gravitational force.

Thus, at the atomic scale, electrical interaction is enormously stronger than gravitational interaction.

1.4 Quantisation of electric charge

(a) Meaning

Electric charge is quantised means that charge exists in integral multiples of the elementary charge $e$.

$$\boxed{q=ne}$$

where

$$n=0,\pm1,\pm2,\pm3,\ldots$$

and

$$e=1.6\times10^{-19}C$$

Thus, a body cannot normally possess an arbitrary fractional value of elementary charge.

(b) Why is quantisation ignored for macroscopic charges?

For a macroscopic body, the number of electrons involved is extremely large.

For example,

$$1C\approx6.25\times10^{18}$$

electrons.

A change of one electron therefore produces an extremely tiny relative change in the total charge.

Hence, macroscopic charge can be treated as continuous.

1.5 Charging of glass rod and silk cloth

When a glass rod is rubbed with silk:

• electrons are transferred from one material to the other.

• The glass rod becomes positively charged.

• The silk becomes negatively charged.

If the glass loses $n$ electrons, its charge becomes

$$+ne$$

and silk receives the same $n$ electrons:

$$-ne$$

Therefore,

$$Q_{\text{total}}=+ne-ne=0$$

Thus, charge is not created or destroyed; it is only transferred from one body to another.

$$\boxed{\text{This is consistent with the law of conservation of charge.}}$$

1.6 Four charges at the corners of a square

Given:

$$q_A=+2\mu C$$ $$q_B=-5\mu C$$ $$q_C=+2\mu C$$ $$q_D=-5\mu C$$

A charge of $1\mu C$ is placed at the centre.

The centre is equidistant from all four corners.

Charges at $A$ and $C$ are equal and opposite in force direction at the centre. Hence their forces cancel.

Similarly, charges at $B$ and $D$ produce equal and opposite forces and cancel.

Therefore,

$$\boxed{F_{\text{net}}=0}$$

Answer:

$$\boxed{0N}$$

1.7 Electric field lines

(a) Why can't a field line have sudden breaks?

Electric field lines represent the direction of the electric field.

They originate from positive charges and terminate on negative charges or at infinity.

A field line cannot suddenly stop in a region where there is no charge because that would imply that the electric field suddenly becomes zero without a physical reason.

Therefore, electric field lines are continuous curves.

(b) Why can two field lines never cross?

At any point, the electric field has a unique direction.

If two field lines crossed, there would be two different tangential directions of the electric field at the same point.

That is impossible.

$$\boxed{\text{Hence, two electric field lines never cross each other.}}$$

1.8 Two opposite point charges

Given:

$$q_A=+3\mu C$$ $$q_B=-3\mu C$$

Distance:

$$AB=20cm=0.20m$$

Midpoint $O$:

$$AO=BO=0.10m$$

(a) Electric field at midpoint

Field due to $q_A$:

$$E_A=\frac{kq}{r^2}$$ $$E_A=\frac{9\times10^9(3\times10^{-6})}{(0.1)^2}$$ $$E_A=2.7\times10^6N/C$$

Similarly,

$$E_B=2.7\times10^6N/C$$

Both fields point from positive charge toward negative charge.

Therefore,

$$E=E_A+E_B$$ $$E=5.4\times10^6N/C$$ $$\boxed{E=5.4\times10^6N/C}$$

Direction: from $A$ toward $B$.

(b) Force on negative test charge

$$q=-1.5\times10^{-9}C$$ $$F=qE$$

Magnitude:

$$F=(1.5\times10^{-9})(5.4\times10^6)$$ $$\boxed{F=8.1\times10^{-3}N}$$

Since the test charge is negative, the force is opposite to the electric field.

Therefore, the force is towards the positive charge $A$.

1.9 Electric dipole

Given:

$$q_A=+2.5\times10^{-7}C$$ $$q_B=-2.5\times10^{-7}C$$

Coordinates:

$$A=(0,0,-15cm)$$ $$B=(0,0,+15cm)$$

Distance:

$$d=30cm=0.30m$$

Total charge

$$Q=q_A+q_B$$ $$Q=2.5\times10^{-7}-2.5\times10^{-7}$$ $$\boxed{Q=0}$$

Electric dipole moment

$$p=qd$$ $$p=(2.5\times10^{-7})(0.30)$$ $$\boxed{p=7.5\times10^{-8}Cm}$$

Direction is from negative charge $B$ to positive charge $A$, i.e. along the negative $z$-axis.

$$\boxed{\vec p=-7.5\times10^{-8}\hat{k}\,Cm}$$

1.10 Torque on an electric dipole

Given:

$$p=4\times10^{-9}Cm$$ $$E=5\times10^4N/C$$ $$\theta=30^\circ$$

Torque:

$$\tau=pE\sin\theta$$ $$\tau=(4\times10^{-9})(5\times10^4)\sin30^\circ$$

Since

$$\sin30^\circ=\frac12$$ $$\tau=4\times10^{-9}\times5\times10^4\times\frac12$$ $$\boxed{\tau=1.0\times10^{-4}Nm}$$

1.11 Polythene rubbed with wool

Given charge on polythene:

$$Q=-3\times10^{-7}C$$

(a) Number of electrons transferred

$$Q=ne$$

Therefore,

$$n=\frac{|Q|}{e}$$ $$n=\frac{3\times10^{-7}}{1.6\times10^{-19}}$$ $$n=1.875\times10^{12}$$ $$\boxed{n\approx1.9\times10^{12}\text{ electrons}}$$

Since polythene becomes negative, it gains electrons.

Therefore, electrons are transferred:

$$\boxed{\text{from wool to polythene}}$$

(b) Is mass transferred?

Yes.

Electrons have mass, so when electrons move from wool to polythene, a very small amount of mass is transferred.

Mass transferred:

$$m=nm_e$$ $$m=(1.875\times10^{12})(9.11\times10^{-31})$$ $$\boxed{m\approx1.71\times10^{-18}kg}$$

So, yes, mass is transferred from wool to polythene, but it is extremely small.

1.12 Two charged copper spheres

(a) Mutual force

Given:

$$q_1=q_2=6.5\times10^{-7}C$$ $$r=50cm=0.50m$$

Using Coulomb's law:

$$F=\frac{kq_1q_2}{r^2}$$ $$F=\frac{9\times10^9(6.5\times10^{-7})^2}{(0.5)^2}$$ $$F\approx1.52\times10^{-2}N$$ $$\boxed{F=1.52\times10^{-2}N}$$

The force is repulsive.

(b) Charges doubled and distance halved

New charge:

$$q'=2q$$

New distance:

$$r'=\frac r2$$

Therefore,

$$F'=\frac{k(2q)^2}{(r/2)^2}$$ $$F'=16F$$

Thus,

$$F'=16(1.52\times10^{-2})$$ $$\boxed{F'\approx0.243N}$$

1.13 Tracks of charged particles in a uniform electric field

This question depends on Figure 1.30, which is not included in your pasted text.

The signs of the three charges and their charge-to-mass ratios are determined from the direction and curvature of the tracks in the figure.

For a charged particle in an electric field:

$$F=qE$$

and

$$a=\frac{qE}{m}$$

Therefore,

$$\boxed{\frac{q}{m}=\frac{a}{E}}$$

The particle whose path shows the greatest acceleration/deflection has the largest magnitude of $q/m$.

I need Figure 1.30 to give the exact signs and identify the particle.

1.14 Electric flux through a square

Given:

$$E=3\times10^3\hat{i}\,N/C$$

Side of square:

$$l=10cm=0.10m$$

Area:

$$A=l^2=(0.10)^2=0.01m^2$$

(a) Plane parallel to yz-plane

The normal to the yz-plane is along the $x$-axis.

Therefore,

$$\theta=0^\circ$$

Electric flux:

$$\Phi=EA\cos\theta$$ $$\Phi=(3\times10^3)(0.01)(1)$$ $$\boxed{\Phi=30Nm^2/C}$$

(b) Normal makes $60^\circ$ with x-axis

$$\Phi=EA\cos60^\circ$$ $$\Phi=(3\times10^3)(0.01)\left(\frac12\right)$$ $$\boxed{\Phi=15Nm^2/C}$$

1.15 Net flux through a cube

A cube is placed in a uniform electric field.

For every face through which flux enters, an equal amount of flux leaves through the opposite face.

Therefore, the total flux is:

$$\boxed{\Phi_{\text{net}}=0}$$

This also follows from Gauss's law because the cube contains no net charge:

$$\Phi=\frac{Q_{\text{inside}}}{\epsilon_0}=0$$

1.16 Net charge inside a black box

Given:

$$\Phi=8.0\times10^3Nm^2/C$$

By Gauss's law:

$$\Phi=\frac{Q}{\epsilon_0}$$

Therefore,

$$Q=\epsilon_0\Phi$$ $$Q=(8.85\times10^{-12})(8.0\times10^3)$$ $$\boxed{Q=7.08\times10^{-8}C}$$

(b) If flux is zero

No.

Zero net flux means:

$$Q_{\text{net}}=0$$

It does not necessarily mean that there are no charges inside.

For example, equal positive and negative charges could be present:

$$(+q)+(-q)=0$$

Thus, the net charge is zero even though charges are present.

$$\boxed{\text{Zero net flux does not imply absence of charges.}}$$

1.17 Flux through a square

Given:

$$q=+10\mu C$$

The square can be considered as one face of a cube.

By symmetry, total flux from the charge is equally distributed among the six faces.

By Gauss's law:

$$\Phi_{\text{total}}=\frac{q}{\epsilon_0}$$

Therefore, flux through one face:

$$\Phi=\frac{q}{6\epsilon_0}$$ $$\Phi=\frac{10\times10^{-6}}{6(8.85\times10^{-12})}$$ $$\boxed{\Phi\approx1.88\times10^5Nm^2/C}$$

1.18 Point charge inside a cubic Gaussian surface

Given:

$$q=2.0\mu C$$

By Gauss's law:

$$\Phi=\frac{q}{\epsilon_0}$$ $$\Phi=\frac{2.0\times10^{-6}}{8.85\times10^{-12}}$$ $$\boxed{\Phi\approx2.26\times10^5Nm^2/C}$$

Notice that the flux is independent of the size or shape of the Gaussian surface, provided the charge remains enclosed.

1.19 Flux from a point charge

Given:

$$\Phi=-1.0\times10^3Nm^2/C$$

(a) Radius doubled

According to Gauss's law:

$$\Phi=\frac{q}{\epsilon_0}$$

The flux depends only on the enclosed charge, not on the radius of the Gaussian sphere.

Therefore,

$$\boxed{\Phi=-1.0\times10^3Nm^2/C}$$

(b) Point charge

$$q=\epsilon_0\Phi$$ $$q=(8.85\times10^{-12})(-1.0\times10^3)$$ $$\boxed{q=-8.85\times10^{-9}C}$$

1.20 Conducting sphere

Given:

$$R=10cm$$

Electric field is measured at:

$$r=20cm=0.20m$$ $$E=1.5\times10^3N/C$$

The field points radially inward, so the charge is negative.

For a charged conducting sphere:

$$E=\frac{k|Q|}{r^2}$$

Therefore,

$$|Q|=\frac{Er^2}{k}$$ $$|Q|=\frac{(1.5\times10^3)(0.20)^2}{9\times10^9}$$ $$|Q|=6.67\times10^{-9}C$$

Since the field is inward:

$$\boxed{Q=-6.67\times10^{-9}C}$$

1.21 Uniformly charged conducting sphere

Given diameter:

$$D=2.4m$$

Therefore,

$$R=1.2m$$

Surface charge density:

$$\sigma=80.0\mu C/m^2$$

(a) Charge on sphere

Surface area:

$$A=4\pi R^2$$ $$A=4\pi(1.2)^2$$ $$A\approx18.10m^2$$

Charge:

$$Q=\sigma A$$ $$Q=(80\times10^{-6})(18.10)$$ $$\boxed{Q\approx1.45\times10^{-3}C}$$

(b) Total electric flux

By Gauss's law:

$$\Phi=\frac{Q}{\epsilon_0}$$ $$\Phi=\frac{1.45\times10^{-3}}{8.85\times10^{-12}}$$ $$\boxed{\Phi\approx1.64\times10^8Nm^2/C}$$

1.22 Infinite line charge

Given:

$$E=9\times10^4N/C$$ $$r=2cm=0.02m$$

For an infinite line charge:

$$E=\frac{\lambda}{2\pi\epsilon_0r}$$

Therefore,

$$\lambda=2\pi\epsilon_0rE$$

Substituting:

$$\lambda=2\pi(8.85\times10^{-12})(0.02)(9\times10^4)$$ $$\boxed{\lambda\approx1.0\times10^{-7}C/m}$$

1.23 Two large parallel metal plates

Surface charge densities are:

$$+\sigma,\quad-\sigma$$

where

$$\sigma=17.0\times10^{-22}C/m^2$$

For an infinite charged sheet:

$$E=\frac{\sigma}{2\epsilon_0}$$

Thus,

$$E=\frac{17.0\times10^{-22}} {2(8.85\times10^{-12})}$$ $$\boxed{E\approx9.6\times10^{-11}N/C}$$

(a) Outer region of first plate

The fields produced by the two plates are equal and opposite outside the plates.

Therefore,

$$\boxed{E=0}$$

(b) Outer region of second plate

Again, the two fields cancel:

$$\boxed{E=0}$$

(c) Between the plates

Between oppositely charged plates, the fields are in the same direction and add:

$$E=\frac{\sigma}{2\epsilon_0} +\frac{\sigma}{2\epsilon_0}$$ $$E=\frac{\sigma}{\epsilon_0}$$ $$E=\frac{17.0\times10^{-22}}{8.85\times10^{-12}}$$ $$\boxed{E\approx1.92\times10^{-10}N/C}$$

Final answers for 1.23

$$\boxed{(a)\;0}$$ $$\boxed{(b)\;0}$$ $$\boxed{(c)\;1.92\times10^{-10}N/C}$$

Quick Answer Summary