Electrostatics

Given
a = 10cm = 0.10m; q =5 μC = 5 X 10^-6 C
.V = Kq / r
Distance of centre from vertex for a regular hexagon is 10 cm.
V= 6 Kq/r
= 6 x 9 x10⁹ x 5 x 10^-6 /0.10
= 2.7 x 10⁶ V
 

C = 10 x 10 ^{- 6} F; V = 100 V; E = ? 
E = ½ C V² = ½  x 10^-5 x 10⁴ =  0.05 J

V = V₁ + V₂ 
Q = 6 x 10^-6 X 2 = 12 μ C
As C₂ is in series, same amount of charge will flow through it.
Now V₂ = Q/C₂ = (12 x 10^-6)/(12 x 10^-6) = 1 V  
Total battery voltage, V = 2 + 1 = 3 V 

The particle will be suspended between the plates, when
E = mg / q
Putting the values in above equation
E = 1 x10^-6 x 9.8 / 1 x10^-6
E = 9.8 NC^-1
 

Three capacitors are connected in series, hence their equivalent capacity will be:
C_s = C / 3 = 5 / 3.
This equivalent series capacity is then connected in parallel with the fourth capacitor, hence the final equivalent capacity will be:
C_s + C ₄ = 5 / 3 + 5 = 20 / 3 = 6.66 6 μF.
Now, charge on C₁, C₂ and C₃ is same as they are in series.
It is given by: Q = C_s V = 5 / 3 x 240 = 5 x 80 x 10 ^{- 6} C = 4 x 10^-4 C
Charge on C₄= C₄ x V = 5 x 240 = 1200 x 10^-6 = 12 x 10^-4C.

Q² = [4πε_o] [Fd²] = [1/9 x 10⁹] x [1.5 x (0.5)²] = 4.17 x 10^-11
Q = 6.45 x 10^-6 C

Distance between the plates -
d = 1cm =1×10^-2 m
We know that,
Strength of electric field (E) = Potential(V )/Distance( d)
E = 3  x 10⁶/10^-2
= 3 x 10⁸ V/m
Which is more than dielectric strength of air, so it is not possible.

Charge = q = 20 e
= 20 x 1.6 x 10^-9
= 32 × 10^-19 C
Potential difference = ΔV = 100 V
We know that:
E = q v
ΔV = (32 × 10^-19) (100)
= 3200 × 10^-19 Joules
Since, 1eV = 1.6 × 10^-19 C
E = 3200 × 10^-19 / 1.6 × 10^-19 = 2000 eV
 

Capacitance of parallel plate capacitor with air between the plates is C₀ = Є₀A/d. 
When the separation between the plates is reduced to half, C₁ = Є₀A/(d / 2) = 2Є₀A/d. 
Thus, final capacitance is C₂ = 10 x 8 pF = 80 pF.

C = 20 x 10^-6 F; Ć = ? 
C = A ε₀K/d
C ∝ 1/d
∴ C = C/2 = 10 μF
C ∝ K
Ć = K C = 20 x 8 = 160 μF

The arrangement is of 5 capacitors in series
Therefore, 1/C’ = (1/C) + (1/C) + (1/C) + (1/C) + (1/C) = (5/C)
 C’ = C/5 Or 5 = C/5 or
C = 25 mF

The electric potential of a charge is given by the equation V = kq/r. In other words, the distance is inversely proportional to electric potential. If the distance is doubled, then the electric potential must be halved.

A small sphere will have a lower potential than a large sphere if they both have the same charge on them. Hence, a current will flow when they are connected.

Applying Gauss’s Law and Putting values E x 4πr² = q / ε₀
or E= q / 4πεor²
This expression is same as electric field due to a point charge q placed at distance r,  i.e. in this
case if complete charge q is placed at the centre of shell the electric field is same. 

L = 0.2 m; b = 0.1 m; d = 2 x 10^-3 m; V = 5 x 102V, E = ?
C = A ε₀/d = (L x b) x ε₀/ d = 2 x 10^-2x 8.85 x 10^-12/2 x 10^-3
∴ C = 8.85 x 10^-11F
Q = C V = 8.85 x 10^-11x 5 x 10² = 4.425 x 10^-8C 
E = V/d = 500 / 2 x 10^-3 = 250 x 10³ V/m

U = 26.55 x 10^-6J/m³; K = 8; E = ?
if U = ½ K ε₀ E²
E² = 2 U/K ε₀ = 2 x 26.55 x 10⁶/8 x 8.85 x 10^-12 ,
E² = 0.75 x 10¹⁸
E = 0.866 x 10⁹ N/C

Energy = ½ CV² = ½Є₀AV²/d = [8.85 x 10^-12 x 25 x 10^-4 x 10⁴] /2 x 0.001 = 1.1 x 10^-7J
New plate separation = 0.002 m; potential across plates is still 100 V.
New energy = 0.5 x original energy = 0.55 x 10^-7 J
The difference in energy is explained by the movement of charge in the wires as the capacitor partly discharges to maintain the potential.