Two identical capacitors C₁ and C₂ get fully charged with 5 V battery initially.
So, the charge and potential difference on both capacitors becomes
q = CV
= 2 x 10^-6 x 5 V = 10\(\mu \)C
and V = 5 V
On introduction of dielectric medium of K = 5.
For C₁ (Continue to be connected with battery)
 Potential difference of C₁ , V' = 5 V
Capacitance C'₁ = KC = 5 x 2 = 10\(\mu \)F
Charge q' = C₁' V' = 10 x 5 = 50\(\mu \)C
For C₂ (Disconnected with battery)
Charge, q' = q = 10\(\mu \)C
\(\therefore\) Potential difference, V' = \(\frac { V }{ K } =\frac{5}{5}=1 V\)

Initially, the charge q₃ is at C. Its distances from q₁ and q₂ are r₁₃ = 80 cm = 0.80 m and r₂₃ =\(\sqrt { { 80 }^{ 2 }+{ 60 }^{ 2 }=100\quad cm } \) 
\(\\ =1.0m\)
Hence, the initial potential energy of the system is
\({ U }_{ i }=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 13 } } +\frac { { q }_{ 2 }{ q }_{ 3 } }{ { r }_{ 23 } } \right] =\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { q }_{ 1 } }{ { r }_{ 13 } } +{ \frac { { q }_{ 2 } }{ { r }_{ 23 } } } \right] \)
\(\\ =\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { 2x10 }^{ -8 } }{ 0.8 } -\frac { { 0.4x10 }^{ -8 } }{ 1.0 } \right]\)
\( \\ =\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } x2.1x{ 10 }^{ -8 }\)
When the charge q₃ is moved to D, its distances from q₂ and q₃ are
r₁₃ = 80 cm = 0.8 m and r₂₃ = 20 cm = 0.2 m
Thus, the final potential energy of the system is
\({ U }_{ f }=\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { { 2x10 }^{ -8 } }{ 0.8 } -\frac { { 0.4x10 }^{ -8 } }{ 0.2 } \right]\)
\( \\ \therefore Decrease\ in\ energy\ of\ the\ system={ U }_{ i }-{ U }_{ f }\)
\(\\ =\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } x(2.1x{ 10 }^{ -8 })-\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } \left( \frac { { 2x10 }^{ -8 } }{ 0.8 } -\frac { { 0.4x10 }^{ -8 } }{ 0.2 } \right)\)
\( \\ =\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } \left[ (2.1x{ 10 }^{ -8 })-\left( \frac { { 2x10 }^{ -8 } }{ 0.8 } -\frac { { 0.4x10 }^{ -8 } }{ 0.2 } \right) \right]\)
\( \\ =\frac { { q }_{ 3 } }{ { 4\pi \varepsilon }_{ 0 } } x1.6x{ 10 }^{ 18 }\)
\(\\ Percentage\ decrease=\frac { { 1.6x10 }^{ -8 } }{ { 2.1x10 }^{ -8 } } =100=76.2%\\ \) 

(a) The point O is equidistant from all the charges at the end points of pentagon.
(b) When charge q is removed from A electric field at O
  E = \(\frac { q\times 1 }{ 4\pi \varepsilon _{ \circ }r^{ 2 } } \)
(c) If charge q at A is replaced by -q then it add chage -2q so electric field become E = \(\frac { 2q }{ 4\pi \varepsilon _{ \circ }r^{ 2 } } \)
(ii) when pentagon is replaced by n-side regular polygon with charge q at each of its corners, the electric field at O would continue to be zero as symmetricity of the charge. 

(i) Charge of proton q = 1.6 x 10^-19C
Force on proton at A is F = qE_A
= (1.6 x 10^-19C) (40 N / C)
= 6.4 x 10^-18N
(ii) Since, electric field,
 E\(\propto \) Number of electric field lines/Area
 Hence, E_B = 1/2 E_A
 = 1/2(40 N / c)
 = 20 N / C.

In the circuit, when initially K ₁ is closed and K₂ is opened, the capacitors C₁ and C₂ acquire potential difference V₁ and V₂, respectively. So we have
V₁ + V₂ = E
and  V₁ + V₂ = 9V
Also, in series combination,\(V \propto 1 / C\) 
V₁ : V₂ = (1/6) : (1/3) 
On solving,
\(\Rightarrow \)  V₁ = 3 V and V₂ = 6V 
\(\therefore \) Q₁ = C₁ V₁  = 6\(\mu\) F x 3 V
= 18\(\mu\) C  [\(\because\) C = 1 \(\mu\)F]
\(\Rightarrow\) Q₂ = C₂V₂
= 3\(\mu\)F x 6V = 18\(\mu\)C 
and   Q₃ = 0
When, K₁ was opened and K₂ was closed, the parallel combination of C₂ and C₃ is in series with C₁ .
[Charge on C₁ remains unchanged]
i.e. Q'₁ = Q₂= \(18\mu C\)  
Charge on C₂ is shared between C₂ and C₃ in parallel.
As, C₂ = C_{3
\(\therefore\) \(Q_{2}^{\prime}=Q_{2}=\frac{Q_{2}}{2}=\frac{18}{2}=9 \mu \mathrm{C}\)} 
 [\(\because\) Q₂ = 18 \(\mu\)C]