Here, \(q_1=2.0\times 10^{-8}C,\)
\(q_2=-1.0\times 10^{-8}C\)
\(C=1.2\times 10^{-3}\mu F=1.2\times 10^{-9}F, V=?\)
\(V={q_1-q_2\over 2C}={{2.0\times 10^{-8}}-(-1.0\times 10^{-8})\over2\times 1.2\times 10^{-9}}\)
=12.5 V

n = 300 turns/m ;
I = 5 A ; l = 0.5 m, r = 10^-2 m.
\({ B }={ \mu }_{ o }nI=\left( 4\pi \times { 10 }^{ -7 } \right) \times 300\times 5\) 
= 1.9 x 10^-3 T.

Here, m = 10^-4kg 
\(q=5\mu C=5\times 10^{-6}C; v=20 ms^{-1}\)
E=\(2\times 10^5 N/C\)
Force on charged particle
\(F=qE=5\times 10^{-6}\times 2\times 10^5=1N\)
As particle is thrown against the field, so 
\(a=-{F\over m}=-{1\over 10^{-4}}=-10^4m/s^2\)
From \(v^2-u^2=2as\)
\((0)^2-(20)^2=2(-10^4)s\)
s = 0.02m

\(Here,\ \lambda =600nm=6\times { 10 }^{ -7 }m,\)
\( D=1.2m,\ n=1,\ x=3mm=3\times { 10 }^{ -3 }m,\ a=?\ \)
\(For \ first \ minimum, \ a \ sin\theta =n\lambda\)
\(a.\left( \frac { x }{ D } \right) =1\lambda\)
\( \\ a=\frac { \lambda D }{ x } =\frac { 6\times { 10 }^{ -7 }\times 1.2 }{ 3\times { 10 }^{ -3 } } =2.4\times { 10 }^{ -4 }m=0.24mm\)

\(Here,\ \Delta m=0.3%(1kg)=\frac { 0.3 }{ 100 } kg\\ =3\times { 10 }^{ -3 }kg\\ E=(\Delta m){ c }^{ 2 }=(3\times { 10 }^{ -3 }){ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }\\ =27\times { 10 }^{ 13 }J\)%(1kg)
\(=\frac { 0.3 }{ 100 } kg\)
\(=3\times { 10 }^{ -3 }kg\)
\(E=(\Delta m){ c }^{ 2 }=(3\times { 10 }^{ -3 }){ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }\)
\(=27\times { 10 }^{ 13 }J\)

Useful Intensity,
\(I=\frac { power }{ area } =\frac { 100\times \left( 2.5/100 \right) }{ 4\pi { \left( 3 \right) }^{ 2 } } =\frac { 2.5 }{ 36\pi } W{ m }^{ -2 }\)
Half of this intensity (I) belongs to electric field and half of that to magnetic field. Therefore,
\(\frac { I }{ 2 } =\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c \ or \ { E }_{ 0 }=\sqrt { \frac { 2I }{ { \varepsilon }_{ 0 }c } } \)
\(=\sqrt { \frac { 2\times \left( 2.5/36\pi \right) }{ \left( \frac { 1 }{ 4\pi \times 9\times { 10 }^{ 9 } } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } =4.08V{ m }^{ -1 }\)
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.08 }{ 3\times { 10 }^{ 8 } } =1.36\times { 10 }^{ -8 }T\)

Applying KVL for loop I, we get
\(-6 I_{1}+12 I_{2}=6\)  .....(i)
Applying KVL for loop II, we get
\( R\left(I_{1}+I_{2}\right)+12 I_{2}=12 \)
\(R I_{1}+(R+12) I_{2}=12\)   ..........(ii)
Putting I₁ = 0, we get \(12 I_{2}=6 \Rightarrow I_{2}=\frac{1}{2} \mathrm{~A}\)
Putting \(I_{2}=\frac{1}{2} \mathrm{~A} \text { and } I_{1}=0 \text { in (ii), }\)
\( (12+R) \frac{1}{2} =12 \)
\(\Rightarrow \ R =12 \Omega\)

\(1.88\times {{10}^{5}}\)Nm²C^-1

(i) -400V
(ii) \(6.4\times {{10}^{-17}} J\)

\(4.5\times {{10}^{3}}V\)

\(4\mu C\)

Energy stored in 6 μF capacitor = E 

\( \therefore \ \frac{1}{2} C_{1} V_{1}^{2}=E \Rightarrow V_{1}^{2}=\frac{2 E}{C_{1}}=\frac{E}{3} \)
\(\because C_{1} \text { and } C_{2} \text { are in parallel, } \therefore V_{1}^{2}=V_{2}^{2}=\frac{E}{3} \)
Now, \(C_{12}=6+12=18 \mu \mathrm{F}\) 
\( \therefore \text { Charge on } C_{12}=q_{12}=C_{12} . V_{12}=18 \times \sqrt{\frac{E}{3}} \)
\(\therefore \text { Charge on } C_{3}=q_{12}=18 \sqrt{\frac{E}{3}} \)
(a) Now, energy stored in 12 μF capacitor
\(=\frac{1}{2} C_{2} V_{1}^{2}=\frac{1}{2} \times 12 \times \frac{E}{3}=2 E\)
(b) Energy stored in 3 μF capacitor
\(=\frac{q_{3}^{2}}{2 C_{3}}=\frac{1}{2} \times\left(\frac{18}{3}\right)^{2} E=18 E\)
\((c) \because C_{e q}=\left(\frac{18 \times 3}{18+3}\right) \mu \mathrm{F}=\frac{18}{7} \mu \mathrm{F}, Q=18 \sqrt{\frac{E}{3}}\)
Energy drawn from the battery,
\(U=\frac{Q^{2}}{2 C_{e q}}=\frac{18 \times 18 \times E \times 7}{3 \times 18}=42 E\)