A small oil drop of mass 10-6 kg is hanging in at rest between two plates separated by 1 mm having a potential difference of 500 V. The charge on the drop is (g = 10 ms-2)

Question

A small oil drop of mass 10-6 kg is hanging in at rest between two plates separated by 1 mm having a potential difference of 500 V. The charge on the drop is (g = 10 ms-2) (A) 2×10-9C (B) 2×10-11C (C) 2×10- 6C (D) 2×10- 8C

Tardigrade Admin · Accepted Answer

Given, the drop rests between the two plates

∴

∴ qE = m g

or

q \frac{V}{r} = m g (∵ E = \frac{V}{r})

\Rightarrow q = \frac{m g r}{V}

here,

m = 1 0^{- 6} k g, g = 10 m / s^{2}, r = 1 mm = 1 0^{- 3} m

and

V = 500 V

Substituting all the values, we get

q = \frac{1 0 ^{- 6} \times 10 \times 1 0 ^{- 3}}{500}, q = 2 \times 1 0^{- 11} C

Q. A small oil drop of mass $1 0^{- 6} k g$ is hanging in at rest between two plates separated by $1 mm$ having a potential difference of $500 V$ . The charge on the drop is $(g = 10 m s^{- 2})$

$2 \times 1 0^{- 9} C$

$2 \times 1 0^{- 11} C$

$2 \times 1 0^{- 6} C$

$2 \times 1 0^{- 8} C$

Q. A small oil drop of mass 10−6kg is hanging in at rest between two plates separated by 1mm having a potential difference of 500V. The charge on the drop is (g=10ms−2)

2×10−9C

2×10−11C

2×10−6C

2×10−8C

Given, the drop rests between the two plates ∴ ∴qE=mg orqrV​=mg(∵E=rV​) ⇒q=Vmgr​ here, m=10−6kg,g=10m/s2,r=1mm=10−3m and V=500V Substituting all the values, we get q=50010−6×10×10−3​,q=2×10−11C

Q. A small oil drop of mass $1 0^{- 6} k g$ is hanging in at rest between two plates separated by $1 mm$ having a potential difference of $500 V$ . The charge on the drop is $(g = 10 m s^{- 2})$

$2 \times 1 0^{- 9} C$

$2 \times 1 0^{- 11} C$

$2 \times 1 0^{- 6} C$

$2 \times 1 0^{- 8} C$