Ans (a).
Dielectric constant of the mica sheet, k= 6

If voltage supply remained connected, voltage between two plates will be constant.

Supply voltage, V = 100 V

Initial capacitance, C= 1.771 x 10^-11 F

New capacitance, C₁ = kC = 6 x 1.771 x 10^-11 F = 106 pF

New charge, q₁ = C₁V = 106 x 100 pC = 1.06 x 10^–8 C

Potential across the plates remains 100 V.

Ans (b).

Dielectric constant, k= 6

Initial capacitance, C = 1.771 x 10^-11 F

New capacitance, C₁= kC = 6 x 1.771 x 10^-11 F = 106 pF

If supply voltage is removed, then there will be constant amount of charge in the plates.

Charge = 1.771 x 10^–9 C

Potential across the plates is given by,

V₁q/C₁= (1.771 x 10^–9)/(106 x 10⁻¹²)

=16.7 V

Ans (a).
Capacitances of the given capacitors: C₁ = 2 pF, C2 = 3 pF and C3 = 4 pF

For the parallel combination of the capacitors, equivalent capacitor is given by C_eq the algebraic sum,

Therefore C_eq = C₁+C2 + C3 = 2 + 3+ 4 = 9pF

Therefore, total capacitance of the combination is 9 pF.

Ans (b).
Supply voltage, V = 100 V

The voltage through all the three capacitors is same = V = 100 V

Charge on a capacitor of capacitance C and potential difference V is given by the relation,

q=VC… (i)

For C = 2 pF, charge = VC = 100 x 2 = 200 pC  = 2   x lO⁻¹⁰ C

For C = 3 pF, charge = VC = 100 x 3 = 300 pC   = 3   x lO⁻¹⁰ C

For C = 4 pF, charge = VC = 100 x 4 = 400 pC  = 4   x 10⁻¹⁰ C

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance between the parallel plates was d and it was filled with air.

Dielectric constant of air, k = 1

Capacitance, C, is given by the formula,

C = (kε₀ A)/d = ε₀ A/d ———————–Eq-1

Where,

A = Area of each plate

ε₀ = Permittivity of free space

If distance between the plates is reduced to half, then new distance, di = d/2 Dielectric constant of the substance filled in between the plates, k₁= 6 Hence, capacitance of the capacitor becomes

C₁ = k₁ ε₀ A/d₁ = (6 ε₀ A)/(d/2) = (12 ε₀ A)/d………………Eq-2

Taking ratios of equations (1) and (2), we obtain C₁ = 2 x 6 C = 12 C = 12 x 8 pF = 96 pF Therefore, the capacitance between the plates is 96 pF.

Velocity of the particle, V_x = 2.0 x 10⁶ m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between the two plates, E = 9.1 x 10² N/C

Charge on an electron, q = 1.6 x 10^–19 C

Mass of an electron, m_e= 9.1 x 10^-31 kg

Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,

s= (qEL²)/(2mV²_x)

=>L = √ (2smV²_x)/qE

=  √ 2 x 0.005 x 9.1 x 10^-31)/(1.6 x 10^–19 x 9.1 x 10²

= √ 0.00025  = 0.016 m=1.6 cm

Therefore ,the electron will strike the upper plate after travelling 1.6cm.

Ans (a).

Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

Ans (b).

Two charges of same magnitude and same sign are placed at a certain distance. The mid­point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.