ELECTROSTATICS

6 ) Free Physics Questions Online Mock Test Series By Cracktest

Two point charges repel each other with a force of 100 N. One of the charges is increased by 10% and other is reduced by 10%. The new force of repulsion at the same distance would be …… N.

View Solution

Answer :

We know F ∝ Q₁Q₂

hence (F' / F) = [(Q₁' Q₂') / (Q₁Q₂)]

Here

F = 100 N

Q₁' → 1.1 Q₁

Q₂' → 0.9 Q₂

∴ [F' / (100)] = [{(1.1 Q₁) (0.9 Q₂)} / (Q₁Q₂)]

∴ F' = 100 × 0.99

F' = 99 N

7 )

Given that q₁ + q₂ = q if the between q₁ and q₂ is maximum, (q₁ / q) ………

0.25

0.5

0.75

View Solution

Answer :

given q₁ + q₂ = q (1)

F = k[(q₁ q₂) / r²]

from (1), F = [{k ∙ q₁(q – q₁)} / r²]

F = (k / r²)(qq₁ – q₁²)

For F is to be maximum, differentiate ωrt q₁

∴ (dF / dq₁) = (k / r²)(q – 2q₁)

For maximum value, (dF / dq₁) = 0

∴ q – 2q₁ = 0

(q₁ / q) = (1 / 2)

8 )

Two small conducting sphere of equal radius have charges + 1 c and – 2c respectively and placed at a distance d from each other experience force F₁. If they are brought in contact and separated to the same distance, they experience force F₂. The ratio of F₁ to F₂ is …………

â€“ 8 : 1

1 : 2

1 : 8

â€“ 2 : 1

View Solution

Answer :

Initial charges : Q₁ = 1c & Q₂ = – 2c.

when identification bodies having charge Q₁ & Q₂ are brought in contact

& separated then charge on each body is [(Q₁ + Q₂) / 2]

hence new charges: Q₁' = Q₂' = [{1 + (– 2)} / 2] = [(– 1) / 2]c

as F ∝ Q₁ Q₂

∴ (F₁ / F₂) = [(Q₁ Q₂) / (Q₁' Q₂')]

∴ (F₁ / F₂) = [{(1) (– 2)} / {[(– 1) / 2] [(– 1) / 2]}]

= [(– 2) / (1/4)]

= – 8

∴ F₁ : F₂ = – 8 : 1

9 )

Two small charged spheres repel each other with a force 2 × 10^–3 N. The charge on one sphere is twice that of the other. When these two spheres displaced 10 cm further apart the force is 5 × 10^–4 N, then the charges on both the spheres are…….

Select correct options from the following options

(A) 1.6 × 10^–9 C, 3.2 × 10^–9 C

(B) 3.4 × 10^–9 C, 11.56 × 10^–9 C

(D) 2.1 × 10^–9 C, 4.41 × 10^–9

View Solution

Answer :

given : F = 2 × 10^–3 N

Let Q₁ = q here Q₂ = 2q

Let r be distance between them

As F = [(kQ₁Q₂) / r₂] hence 2 × 10^–3 = [{k(q) (2q)} / r²]

∴ 2 × 10^–3 = [(2q²k) / r²]

∴ [(kq²) / r²] = 10^–3 (1)

when r is changes to (r + 0.1)m, force is 5 × 10^–4 N

∴ 5 × 10^–4 = [{k(q) (2q)} / (r + 0.1)²]

∴ 5 × 10^–4 = [(2q²k) / (r + 0.1)²]

from (1) q²k = r² × 10^–3

∴ 5 × 10^–4 = [(2 × 10^–3 × r²) / (r + 0.1)²]

∴ 2.5 × 10^–1 = [r / (r + 0.1)]²

∴ 0.25 = [r / (r + 0.1)]²

∴ 0.5 = [r / (r + 0.1)]

0.5r + 0.05 = r

∴ 0.05 = 0.5r

r = 0.1 m

from (1) [(9 × 10⁹ × q²) / (0.1)²] = 10^–3

∴ q² = [(10^–3 × 10^–2) / (9 × 10⁹)] = [(10^–14) / 9]

q = [(10^–7) / 3] = 3.33 × 10^–8 C

& 2q = (2/3) × 10^–7 = 6.66 × 10^–8 C

∴ charges are 33.33 × 10^–9 C & 66.66 × 10^–9 C

10 )

Two point charges of + 16 c and – 9 c are placed 8 cm apart in air ……. distance of a point from – 9 c charge at which the resultant electric field is zero.

View Solution

Answer :

Let that point is at distance x from – 9 c charge

[(kq₁) / (x + 0.08)²] = [(kq²) / x²]

Here q₁ = 16 c, q₂ = – 9 c

∴ [(16) / (x + 0.08)²] = (9/x²)

∴ [x / (x + 0.08)]² = (9 / 16)

[x / (x + 0.8)] = (3/4)

4x = 3x + 0.24

x = 0.24m = 24cm

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