ELECTROSTATICS

11 )

A small sphere whose mass is 0.1 gm carries a charge of 3 × 10^–10C and is tie up to one end of a silk fiber 5 cm long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of 25 × 10^–6Cm^–2, on each side. When system is freely hanging the angle fiber makes with vertical is ………

41.8Â°

45Â°

40.2Â°

45.8Â°

View Solution

Answer : Free Physics Questions Online Mock Test Series By Cracktest

m = 0.1 × 10^–3 kg

q = 3 × 10^–10C

ρ = surface charge density = 25 × 10^–6 C/m²

from figure : T sin θ = qE

T cos θ = mg

∴ Taking ratio, we get tan θ = (qE / mg) (1)

Also E = (ρ / ∈₀) = [(25 × 10^–6) / (8.85 × 10^–12)] = 2.82 × 10⁶

∴ from (1),

tan θ = [(3 × 10^–10 × 2.82 × 10⁶) / (0.1 × 10^–3 × 10)]

= 8.47 × 10^–4 × 10³

tan θ = 0.847

θ = tan^–1 (0.847) = 40.2°

12 )

Two uniformly charged spherical conductors A and B having radius 1mm and 2mm are separated by a distance of 5 cm. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is…..

4 : 1

1 : 2

2 : 1

1 : 4

View Solution

Answer :

In equilibrium condition, potential of sphere are equal.

if potentials are V_A & V_B then V_A = V_B

As V = E ∙ R

i.e. E = (V/R)

∴ (E_A / E_B) = [(V_A / R_A) / (V_B / R_A)] = (V_A / V_B) / (R_B / R_A)] = 1 ∙ (2/1) = (2/1)

13 )

A simple pendulum consists of a small sphere of mass m suspended by a thread of length ℓ. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is

Select correct option from the following options

(A) T = 2π[ℓ / {g – (qE / m)}]^(1/2)

(B) T = 2π(ℓ/g)^(1/2)

(D) T = 2π[(mℓ / qE)]^(1/2)

View Solution

Answer :

The sphere of mass m has charge + q.

The electric field E is directed vertically up.

hence sphere will experience force in upward direction.

This forces will be F = qE

Due to this force, acceleration produced in the upward direction in sphere

is a = (F/m) = (qE / m)

The net acceleration on sphere is g' = g – a = g – (qE / m)

As = Time period = T = 2π√(ℓ / g')

∴ T = 2π√[ℓ / {g – (qE / m)}]

14 )

In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. 2400 v between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 v. What is the charge on the second drop?

[(3Q) / 2]

(Q / 4)

(Q / 2)

View Solution

Answer :

F = qE also F = mg

in balanced condition, qE = mg (1)

as E = Electric field = [(Potential) / (distance)] & ρ = [(mass) / volume)]

∴ E = (v / d) & m = ρ ∙ v

from (1)

(qv / d) = ρ ∙ (4/3)πr³g

q = (4/3)πgρd(r³ / v)

q ∝(r³ / v) ------ other quantities are constant

∴ (q₁ / q₂) = (r₁ / r₂)³ × (v₂ / v₁)

= [r₁ / (r₁ / 2)]³ × [(600) / (2400)]

= 8 × (1/4) = 2

∴ q₂ = (q₁ / 2)

∴ q₂ = (Q / 2)

15 )

A charged particle of mass 1 kg and charge 2μc is thrown from a horizontal ground at an angle θ = 45° with speed 20m/s. In space a horizontal electric field E = 2 × 10⁷ V/m exist. The range on horizontal ground of the projectile thrown is ………..

100 m

200 m

150 m

50 m

View Solution

Answer :

E = 2 × 10⁷ V/m

q = 2μc

m = 1 kg

V = 20 m/s

θ = 45°

Time of flight of projectile is

T = [(2V sin θ) / g] = [(2 × 20 sin 45) / (10)]

T = 2√2 sec

Horizontal range of projection is

x = V_xT + (1/2)a_xT²

= (V cos 45)T + (1/2)a_xT²

Now a_x = (qE / m) = [(2 × 10⁷ × 2 × 10^–6) / 1] = 40 m/s

∴ x = (20 cos 45) × 2√2 + [(1/2) × 40 × (2√2)²}]

= [(20 / √2) × 2√2] + [20 × (4 × 2)]

= 40 + 160

= 200 m

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