(a) Acceleration of the electron \(a=\frac{F}{m}\)
Electrostatic force F = - eE
\(a=\frac{F}{m}=\frac{-e E \hat{j}}{m}\)   ... ( 1)
(b) Let velocity of the electron be \(\vec{v}\)
v = u + at        ...(2)
Here, \( u=v_{0} \hat{i} \), \(a=\frac{-e \bar{E}}{m} \hat{j}\)
Substituting these values in the equation (2) we get
\(\therefore \vec{v}=\nu_{0} \hat{i}-\frac{e E}{m} t \hat{j}\)
(c) Displacement \(\vec{r}\) represents position,
\( s=u t+\frac{1}{2} a t^{2} \) ...(3)
Let, \(s=\hat{r} \) and Here, \(u=v_{0} \hat{i} \), \(a=\frac{-e E}{m} \hat{j}\)
Substituting these values in the equation (3), we get,
\(\hat{r}=v_{0} t \hat{i}-\frac{1}{2} \frac{e E}{m} t^{2} \hat{j}\)

\(W =\left(V_{b}-V_{a}\right) q\left[where\ V=\frac{K Q}{r}\right] \)
To find V_b :
\(V_b=\frac{kQ}{r_3}+\frac{kQ}{r_4}\)
\(V_{b} =\frac{K \times 10 \times 10^{-6}}{\sqrt{50 \times 10^{-4}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{250 \times 10^{-4}}} \)
\(V_{b} =\frac{K \times 10^{-5}}{\sqrt{50} \times 10^{-2}}+\frac{K \times 10^{-5}}{\sqrt{250} \times 10^{-2}} \)
\(=K \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
\(=9 \times 10^{9} \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
V_b = 1842002 V
To find V_a :
\(V_a=\frac{kQ}{r_1}+\frac{kQ}{r_2}\)
\(V_{a} =\frac{K \times 10 \times 10^{-6}}{\sqrt{5 \times 10^{-2}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{15 \times 10^{-2}}} \)
V_a = 2400000 V
\(\therefore W_{D} =\left(V_{b}-V_{a}\right) q=(1842002-2400000)( -2 \times 10^{-6} )\)
W = +1.12J
Positive sign implies that to move the charge - 2 μC external work is required.

C_p = C_b + C_c
\(C_{P}=6+2=8 \mu \mathrm{F} \)
\(\frac{1}{C_{s}}=\frac{1}{C_a}+\frac{1}{C_p}=\frac{1}{C_d} \)
\(C_{s}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8} \)
\(\therefore C_{s}=\frac{8}{3} \mu \mathrm{F} \)
Total capacitance \(C_{s}=\frac{8}{3} \times 10^{-6} \mathrm{~F} \)
Total Charge, \(Q=C_{s} V=\frac{8}{3} \times 10^{-6} \times 9 \)
\(Q_{a}=24 \mu C\)
(a) Charge on capacity \(Q_{a}=24 \mu C\)         .....(1)
Charge on capacitor \(Q_{b}=24 \times \frac{6}{8}=18 \mu \mathrm{C} \)             ..............(2)
Charge on capacitor  \(Q_{c}=24 \times \frac{2}{8}=6 \mu \mathrm{C} \)              ..............(3)
Charge on capacitor \(Q_{d}=24 \times \frac{8}{8}=24 \mu \mathrm{C} \)              ..............(4)
(b) Potential difference across \(C_{a} \ is\ V_{a}=\frac{Q_{a}}{C_{a}} \)
\(=\frac{24}{8}=3 \mathrm{~V} \)      ...(5)
Potential difference across \(C_{b}\ is \ V_{b}=\frac{Q_{b}}{C_{b}} \)
\(=\frac{18}{6}=3 \mathbf{V}\)       ........(6)
Potential difference across \(C_{c}\ is \ V_{c}=\frac{Q_{c}}{C_{c}}=\frac{6}{2}=3 \mathrm{~V} \)       ......(7)
Potential difference across \(C_{d}\ is \ V_{d}=\frac{Q_{d}}{C_{d}} \)
\(=\frac{24}{8}=3 \mathbf{V} \)          ....(8)
(c) Energy stored in each capacitor \(U=\frac{1}{2} C V^{2}\)
Energy stored in \(\mathrm{C}_{\mathrm{a}} \text { is } U_{a}=\frac{1}{2} C_{a} V_{a}^{2}\)
\(U_{\mathrm{a}}=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3=36 \mu \mathrm{J}\)   ....(9)
Energy stored in \(C_{b}\ is \ U_{b}=\frac{1}{2} C_{b} V_{b}^{2} \)
\(U=\frac{1}{2} \times 6 \times 10^{-6} \times 3 \times 3 \)
\(=27 \mu \mathrm{J} \)         ........(10)
Energy stored in C_e is \(U_{c} =\frac{1}{2} C_{c} V_{c}^{2} \)
\(=\frac{1}{2} \times 2 \times 10^{-6} \times 3 \times 3=9 \mu \mathrm{J} \)          ......(11)
Energy stored in C_d is \(U_{d} =\frac{1}{2} C_{d} V_{d}^{2} \)
\(U_d=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3 \)
\(=36 \times 10^{-6} \mathrm{~J}=36 \mu \mathrm{J} \)      .........(12)

(i) Electric field is given by \(|E|=\frac{d V}{d x}\)
(a) For region A,
dV = 8 - 5 = 3V, dx - (0.2 - 0) = 0.2m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{3}{0.2}=15 \mathrm{Vm}^{-1}\)
(b) For region B,
dv = 0, E_x = 0
(c) For region C
dV = (7 - 5) = 2V, dx = (0.6 - 0.4) = 0.2 m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{2}{0.2}=10 \mathrm{Vm}^{-1}\)
(d) For region D
dV = (7 - 1)=6V,  dx = (0.8 - 0.6) = 0.2 m
\(\therefore E_{x}=\frac{d V}{d x}=\frac{6}{0.2}=30 \mathrm{Vm}^{-1}\)
(ii) Electric field as a function of x for Figure (ii).

h = 1 x 10^-3 m, e = 1.6 x 10^-19 C, m_e = 9.1 x 10^-31 kg, m_p =1.6 x 10^-27 kg
(a) (i) Time of flight for electron \(t_{c}=\sqrt{\frac{2 h m_{e}}{e E}}\)
Where,
\(E =\frac{V}{d}=\frac{5}{1 \times 10^{-3}}=5000 \mathrm{Vm}^{-1} \)
\(t_{e} =\sqrt{\frac{2 \times 1 \times 10^{-3} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 5000}}=\sqrt{\frac{18.2 \times 10^{-34}}{8 \times 10^{-16}}}=\sqrt{2.275 \times 10^{-18}} \)
= 1.5 x 10^-9s
t_e = 1.5 ns (ignoring the gravity)
(ii) Time of flight for proton \(t_{p}=\sqrt{\frac{2 h m_{p}}{e E}}\)
\(t_{p} =\sqrt{\frac{2 \times 1 \times 10^{-3} \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times 5000}} \)
\(=\sqrt{\frac{3.2 \times 10^{-30}}{8 \times 10^{-16}}}=\sqrt{\frac{3.2}{8} \times 10^{-14}}=\sqrt{4000 \times 10^{-18}} \)
t_p = 63 x 10^-9s
t_p = 63 ns (ignoring the gravity)
(b) Time of flight for neutron, \(t_{n}=\sqrt{\frac{2 h}{g}}\)
\(t_{n} =\sqrt{\frac{2 \times 1 \times 10^{-3}}{10}}=\sqrt{2 \times 10^{-4}} \)
\(=1,414 \times 10^{-2} \)
\(=14.14 \times 10^{-3} \mathrm{~s} \)
t_n = 14.14 ms
(c) The value of Time period for electron is the least. So, electron will reach the bottom first.