ELECTROSTATICS

1 ) Free Physics Questions Online Mock Test Series By Cracktest

When a Piece of Polythene is rubbed with wool, a charge of – 2 × 10^–7 is developed on polythene. The mass transferred to polythene is …. kg.

Select correct option from following options

(A) 11.38 × 10^–19 (B) 5.69 × 10^–19

View Solution

Answer : Free Physics Questions Online Mock Test Series By Cracktest

Here number of electrons are transferred from wool to polythene.

amount of charge on polythene = – 2 × 10^–7 = q

amount of charge on electron = – 1.6 × 10^–19 = e

Let number of electrons transferred be n.

∴ n = (q/e) = [(– 2 × 10^–7) / (– 1.6 × 10^–19)]

∴ n = 1.25 × 10^–12

mass of e = m_e = 9.1 × 10^–31 kg.

hence mass transferred = n ∙ m_e

= 9.1 × 10^–31 × 1.21 × 10¹²

= 11.38 × 10^–19 kg.

2 )

The protonic charge in 100 gm of water is ……. c

4.8 Ã— 10^5

5.4 Ã— 10^6

3.6 Ã— 10^4

4.9 Ã— 10^6

View Solution

Answer :

density of water = 1000 kg/m³

atomic weight = 18

∴ no of atoms in 18 gm = 6.023 × 10²³ ---------- (Avogadro's numbers)

each molecule of H₂O contains10 protons (8 → O₂, 2 → H₂)

∴ No of protons in 100 gm = [{(6.023 × 10²³) / (18)} × 100] × 10

= 3.3 × 10²⁵

Hence protonic charge = 3.3 × 10²⁵ × 1.6 × 10^–19

= 5.4 × 10⁶ c

3 )

A copper sphere of mass 2 gm contains about 2 × 10²² atoms. The charge on the nucleus of each atom is29e. What fraction of electrons removed from the sphere to give it a charge of 2μc?

2 Ã— 10^â€“10

1.19 Ã— 10^â€“12

1.25 Ã— 10^â€“11

2.16 Ã— 10^â€“11

View Solution

Answer :

Charge of each nucleus = 29e

∴ Net charge 2 gm sphere = (29e) × (2 × 10²²) = 5.8 × 10²³e c.

∴ No of electrons on sphere = 5.8 × 10²³

∴ No of electrons removed to give 2μc charge = (q/e)

= [(2 × 10^–6) / (1.6 × 10^–19)]

= 1.25 × 10¹³

∴ fraction of electrons removed = [(1.25 × 10¹³) / (5.8 × 10²³)]

= 2.16 × 10^–11

4 )

The rate of alpha particle falls on neutral sphere is 10¹² per second. The time in which sphere gets charged by 2μc is ….…… sec

2.25

3.15

6.25

1.66

View Solution

Answer :

An alpha particle carries 2e charge

i.e. charge carried = 2 × 1.6 × 10^–19 c = 3.2 × 10^–19 c.

rate of ∝ particles = 10¹² per second

i.e. in 1 second 10¹² ∝ particles are falling.

∴ charge carried per second = 10¹² × 3.2 × 10^–19 = 3.2 × 10^–7 c

Hence to get 2μc charge, time required will be

t = [(2 × 10^–6) / (3.2 × 10^–7)] = 6.25 sec

5 )

A charge Q is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are……

(Q/3), (Q/3)

(Q/2), (Q/2)

(Q/4), (3Q / 4)

(Q/5), (4Q / 5)

View Solution

Answer :

F ∝ Q₁ Q₂

considering options –

(A) Q₁ Q₂ = (Q/3) (Q/3) = (Q² / 9)

(B) Q₁ Q₂ = (Q/2) (Q/2) = (Q² / 4)

(D) Q₁ Q₂ = (Q/5) (4Q / 5) = (4Q² / 25)

hence F is maximum for charge (Q/2) & (Q/2)

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