Here , q = 25 x10^-9C, 2a =6 m, r=4 m,
p=q(2a)= 25x10^-9x 6  = 1.5 x10^-7 C-m
i) \( E_{\text {axil }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 p r}{\left(r^2-a^2\right)^2} \)i
Now, \(=\frac{9 \times 10^9 \times 2 \times 1.5 \times 10^{-7} \times 4}{\left(4^2-3^2\right)^2}=\frac{2700 \times 4}{49} \)
\(=220.4 \mathrm{NC}^{-1} \)
ii)\( \therefore E_{\text {equitorial }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{\left(r^2+a^2\right)^{3 / 2}} =\frac{9 \times 10^9 \times 1.5 \times 10^{-7}}{\left(4^2+3^2\right)^{3 / 2}}=\frac{1350}{125}=10.8 \mathrm{~N} \mathrm{C}^{-1} \)

As we know that, the number of field lines entering in the cube is the same as that the number of field lines leaving the cube So, no flux is remained on the cube and hence, the net flux over the cube is zero.

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.

Here, E = 9 \(\times\)10⁴ N/C, r = 2 cm = 2 \(\times\)10^-2 m and \(\lambda\)= ?
As, \(E=\frac{\lambda}{2\pi \varepsilon _{0}r}\Rightarrow \lambda =2\pi \varepsilon _{0}rE\)
\(=\frac{1}{2\times 9\times 10^{9}}\times 2\times 10^{-2}\times 9\times 10^{4}=10^{-7}Cm^{-1}\)

Let the value of unknown charge be q.
Electric field at 20 cm away, E = 1.5 \(\times\) 10³ N/C

From the formula, electric field,
\(\begin{aligned} E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \end{aligned}\)
\(\begin{aligned} \Rightarrow 1.5 \times 10^3 & =\frac{9 \times 10^9 \times \mathrm{q}}{\left(20 \times 10^{-2}\right)^2} \\ \end{aligned}\)
\(\begin{aligned} \therefore \quad q =\frac{1.5 \times 10^3 \times 20 \times 20 \times 10^{-4}}{9 \times 10^9} \\ \end{aligned}\)
\(\begin{aligned} & =6.67 \times 10^{-9} \mathrm{C} \end{aligned}\)
As the electric field is radially inwards which shows that the nature of unknown charge q is negative
\(\begin{aligned} & \mathrm{E}_i=\lim _{q_0 \rightarrow 0} \frac{\mathrm{F}_i}{q_0}=\lim _{q_0 \rightarrow 0}\left[\frac{1}{q_0}\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i q_0}{r_i^2} \hat{\mathrm{r}}_i\right)\right] \\ \end{aligned}\)
\(\begin{aligned} & \mathrm{E}_i=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i}{r_i^2} \hat{\mathrm{r}}_i \end{aligned}\)
If E is electric field at point P due to the system of charges, then by principle of superposition of electric fields, E = E₁ + E₂ + E₃ + ... + E_n = \( \sum_{i=1}^{n}E_{i}\)
Using Eq. (i), we get \(\mathrm{E}=\sum_{i=1}^n \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_i}{r_i^2} \hat{\mathbf{r}}_i\)
or \(\mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{\mathbf{r}}_i\)
\(\therefore\) E is a vector quantity.

(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.