Net force on charge q₁ is zero as electric field inside a charged conducting shell is zero.
(ii) Due to induction charge inside the shell is -q₁ and outside the shell net charge will be (Q + q₁)
Electric field due to charged sphere at point
P is \(\mathrm{E}_{\mathrm{P}}=\frac{k\left(\mathrm{Q}+q_{1}\right)}{x^{2}}\)
Force on charge q₂ \(=q_{2} \mathrm{E}=\frac{k\left(\mathrm{Q}+q_{1}\right) \cdot q_{2}}{x^{2}}\)
(iii) Force on spherical shell has same magnitude
\(=\frac{k\left(\mathrm{Q}+q_{1}\right) q_{2}}{x^{2}}\)
(iv) \(\text { Charge density }=\frac{\text { Total charge on the sphere }}{\text { Surface area of sphere }} \)
\(\therefore \sigma=\frac{\left(Q+q_{1}\right)}{4-R^{2}} \)
(v) Net electric flux through the sphere is given
\(\phi=\frac{q_{1}}{\varepsilon_{0}^{2}}\)               [by Gauss's Theorem]

(i) \(\vec{\tau}=\vec{p} \times \vec{E}\)
or \(\tau=p E \sin \theta\)
here p = 2aq
(If point charges are q and -q separated by a distance 2a.)
(ii) Torque is perpendicular to dipole moment and electric field. \(\vec{\tau} \perp \vec{p} \text { and } \vec{\tau} \perp \vec{E}\).
(iii) (a) Maximum Torque \(\tau\) = pE when \(\theta\) = 90°

(b) \(\tau=\frac{p E}{2} \text { when, } \sin \theta=\frac{1}{2} \text { i.e., } \theta=30^{\circ} \text { or } 150^{\circ}\)
​​​​​​
\(\therefore \theta\) = 0° or 180°
\(\therefore \tau \) = pE sin 0° = 0
\(\therefore \tau\) = minimum

(i) \(\frac{q_{1}}{q_{2}}=\frac{8}{4}=\frac{2}{1}\) [Count the Number of field lines associated with each charge]
(ii) \(\because\) q₂ < q₁ and q₁ is positive and q₂ is negative, \(\therefore\)Null point will lie at point C to the right side of point B.
Reason: Electric field at C is in opposite direction as shown.

(iii)  \(\therefore\) Net electric field at C = 0
\(\therefore \quad\left|\vec{E}_{1}\right|=\left|\vec{E}_{2}\right|\)
\(\frac{k\left|q_{1}\right|}{(\mathrm{AC})^{2}}=\frac{k\left|q_{2}\right|}{(\mathrm{BC})^{2}}\)
Here AB = 10 cm = 0.1 m
Let BC = x
\(\frac{q_{1}}{q_{2}}=\left(\frac{\mathrm{AC}}{\mathrm{BC}}\right)^{2} \Rightarrow \frac{2}{1}=\left(\frac{0.1+x}{x}\right)^{2}\)
Taking square root
\(\sqrt{2}=\frac{0.1+x}{x} \quad\left[\because \sqrt{\frac{q_{1}}{q_{2}}}=\sqrt{\frac{2}{1}}\right]\)
\((\sqrt{2}-1) x=0.1\)
\(x=\frac{0.1}{(\sqrt{2}-1)}=\frac{0.1}{\sqrt{2}-11} \times \frac{(\sqrt{2}+1)}{\sqrt{2} \mp 1}\)
x = 0.141 + 0.1 = 0.241 m
or x = 24.1 cm from point B.
(iv) No. As the electric field lines are curved, and at any point on it a test charge (+ve) will experience acceleration but direction of acceleration and velocity may not be same

(i) (a): Gauss's law is applicable for any closed surface. Gauss's law is most useful in situation where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over the plane.
Whereas electric dipole is a system of two equal and opposite point charges separated by a very small and finite distance.
So both statements are correct.
(ii) (b): According to Gauss's law, the electric flux through the sphere is 
\(\phi=\frac{q_{\mathrm{in}}}{\varepsilon_{0}}=\frac{8.85 \times 10^{-13} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=0.1 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}\)
(iii) (c) : For uniformly volume charge density,
\(E=\frac{\rho r}{3 \varepsilon_{0}}\)
\(E \propto r\)
(iv) (a): r = 25 ern = 0.25 m \(\sigma=\frac{3}{\pi} \mathrm{C} / \mathrm{m}^{2}\)
As, \(\sigma=\frac{q}{4 \pi r^{2}} \Rightarrow q=4 \pi \times(0.25)^{2} \times \frac{3}{\pi}=0.75 \mathrm{C}\)
(v) (b): The line charge density at a point on a line is the charge per unit length of the line at that point
\(\lambda=\frac{d q}{d L}\)
Thus, the SI unit for \(\lambda \text { is } \mathrm{Cm}^{-1} \text {. }\)

(I) (d): If there is only one type of charge in the universe then it will produce electric field somehow. Hence Gauss's law is valid.
(ii) (c)
(iii) (c) : According to Gauss's theorem,
Electric flux through the sphere = \(\frac{q}{\varepsilon_{0}}\)
\(\therefore\) Electric flux through the hemisphere = \(\frac{1}{2} \frac{q}{\varepsilon_{0}}\)
= \(\frac{10 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}}=0.56 \times 10^{6} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}\)
\(\approx 0.6 \times 10^{6} \mathrm{Nm}^{2} \mathrm{C}^{-1}=6 \times 10^{5} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1}\)
(iv) (d): As flux is the total number of lines passing through the surface, for a given charge, it is always the charge enclosed \(Q / \varepsilon_{0}\).If area is doubled, the flux remains the same.
(v) (d): As net charge on a dipole is \((-q+q)=0\)
Thus, when a gaussian surface encloses a dipole, as per Gauss's theorem, electric flux through the surface,
\(\oint \vec{E} \cdot d \vec{S}=\frac{q}{\varepsilon_{0}}=0\)

(i) (a): As, \(q E=m g \Rightarrow q=\frac{1.08 \times 10^{-14} \times 9.8}{1.68 \times 10^{5}}\)
\(=6.4 \times 10^{-19} \mathrm{C}\)
(ii) (a): \(q=n e \text { or } \Rightarrow n=\frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}}=4\)
(iii) (c) : For the drop to be stationary,
Force on the drop due to electric field = Weight of the drop
qE=mg
\(q=\frac{m g}{E}=\frac{1.6 \times 10^{-6} \times 10}{100}=1.6 \times 10^{-7} \mathrm{C}\)
Number of electrons carried by the drop is
\(n=\frac{q}{e}=\frac{1.6 \times 10^{-7} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=10^{12}\)
(iv) (c)
(v) (d): Millikan's experiment confirmed that the charges are quantized, i.e., charges are small integer multiples of the base value which is charge on electron. The charges on the drops are found to be multiple of 4. Hence, the quanta of charge is 4 \(\mu \)C.

(i) (a): From Newton's law
\(F=m \vec{a} \text { or } q E=m \vec{a} \Rightarrow a=\frac{q E}{m}=\frac{e E}{m}\)
Using, \(s=u t+\frac{1}{2} a t^{2}\)
\(\therefore \quad h=0+\frac{1}{2} \times \frac{e E}{m} t^{2} \Rightarrow t=\sqrt{\frac{2 h m}{e E}}\)
(ii) (c)
(iii) (b): Force is same in magnitude for both.
\(\therefore \quad m_{1} a_{1}=m_{2} a_{2}\)
\(\frac{a_{1}}{b_{2}}=\frac{m_{2}}{m_{1}}=\frac{1}{0.5}=2\)
(iv) (b): Here \(u=0 ; a=\frac{q E}{m} ; s=y\)
Using, \(v^{2}-u^{2}=2 a s \Rightarrow v^{2}=2 \frac{q E}{m} y\)
\(\therefore \quad \mathrm{K.E.}=\frac{1}{2} m v^{2}=q E y\)
(v) (b): If charge particle is put at rest in electric field, then it will move along line of force.

(i) (d): Electric flux \(\phi=\vec{E} \cdot \vec{A}=E A \cos \theta\)
where \(\vec{A}=A \hat{n}\)
For electric flux passing through \(S_{6}, \hat{n}_{S_{6}}=-\hat{i} \text { (Back) }\)
\(\therefore \quad \phi_{S_{6}}=-\left(4 \times 10^{3} \mathrm{NC}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 37^{\circ}\)
= -32 N m² C-I
(ii) (a) : For electric flux passing through S_1,
\(\hat{n}_{S_{1}}=-\hat{j} \text { (Left) }\)
\(\therefore \quad \phi_{S_{1}}=-\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos \left(90^{\circ}-37^{\circ}\right)\)
= -24 N m² C^-1
(iii) (c): Here, \(\hat{n}_{S_{2}}=+\hat{k}(\text { Top })\)
\(\therefore \quad \phi_{S_{2}}=-\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 90^{\circ}=0\)
\(\hat{n}_{S_{3}}=+\hat{j} \text { (Right) }\)
\(\hat{n}_{S_{4}}=-\hat{k}(\text { Bottom })\)
\(\therefore \quad \phi_{S_{4}}=\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 90^{\circ}=0\)
And, \(\tilde{n}_{S_{5}}=+\hat{i} \text { (Front) }\)
\(\therefore \quad \phi_{S_{5}}=+\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 37^{\circ}\)
= 32 N m² C^-1
S₂ and S₄ surface have zero flux.
(iv) (d): As the field is uniform, the total flux through the cube must be zero, i.e., any flux entering the cube must leave it.
(v) (b): Surface integral \(\oint \vec{E} \cdot d \vec{S}\)is the net electric flux over a closed surface S.
\(\therefore \quad\left[\phi_{E}\right]=\left[\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]\)

(i) (b): The proportionality constant k depends on the nature of the medium between the two charges.
(ii) (c): \({ As, }\left[\varepsilon_{0}\right]=\frac{1}{4 \pi F} \cdot \frac{q_{1} q_{2}}{r^{2}} =\frac{[\mathrm{AT}]^{2}}{\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]\left[\mathrm{L}^{2}\right]} \)
\(=\left\lfloor\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)
(iii) (b)
(iv) (d): \(F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}}\)
\(\therefore\left(10 \times 10^{-3}\right) \times 10=\frac{\left(9 \times 10^{9}\right) \times q^{2}}{(0.6)^{2}}\)
\(\text { or } \ q^{2}=\frac{10^{-1} \times 0.36}{9 \times 10^{9}}=4 \times 10^{-12}\)
\(\text { or } \ q=2 \times 10^{-6} \mathrm{C}=2 \mu \mathrm{C}\)
(v) (b)