A charged oil drop of mass 2.5 × 10-7 kg is in space between the two piates, each of area 2 × 10-2 m2 of a parallel piate capacitor. When the upper plate has a charge of 5 × 10-7 C and the lower plate has an equal negative charge, the oil remains stationary. The charge of the oil drop is (Take g = 10 m/s2 )

Question

A charged oil drop of mass 2.5 × 10-7 kg is in space between the two piates, each of area 2 × 10-2 m2 of a parallel piate capacitor. When the upper plate has a charge of 5 × 10-7 C and the lower plate has an equal negative charge, the oil remains stationary. The charge of the oil drop is (Take g = 10 m/s2 ) (A) 9 × 10-1 C (B) 9 × 10-6 C (C) 8.85 × 10-13 C (D) 1.8 × 10-14 C

Tardigrade Admin · Accepted Answer

We know that qE = mg (qQ/ε0 A) = mg or q = (ε0 A mg/Q) = (8.85 × 10-12 × 2 × 10-2 × 2.5 × 10-7 × 10/5 × 10-7) = 8.85 × 10-13 C

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

$9 \times 1 0^{- 6} C$

$8.85 \times 1 0^{- 13} C$

$1.8 \times 1 0^{- 14} C$

We know that
$qE = m g$
$\frac{qQ}{ε _{0} A} = m g or q = \frac{ε _{0} A m g}{Q}$
$= \frac{8.85 \times 1 0 ^{- 12} \times 2 \times 1 0 ^{- 2} \times 2.5 \times 1 0 ^{- 7} \times 10}{5 \times 1 0 ^{- 7}}$
$= 8.85 \times 1 0^{- 13} C$

9×10−1C

9×10−6C

8.85×10−13C

1.8×10−14C

We know that qE=mg ε0​AqQ​=mgorq=Qε0​Amg​ =5×10−78.85×10−12×2×10−2×2.5×10−7×10​ =8.85×10−13C

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

$9 \times 1 0^{- 6} C$

$8.85 \times 1 0^{- 13} C$

$1.8 \times 1 0^{- 14} C$

We know that
$qE = m g$
$\frac{qQ}{ε _{0} A} = m g or q = \frac{ε _{0} A m g}{Q}$
$= \frac{8.85 \times 1 0 ^{- 12} \times 2 \times 1 0 ^{- 2} \times 2.5 \times 1 0 ^{- 7} \times 10}{5 \times 1 0 ^{- 7}}$
$= 8.85 \times 1 0^{- 13} C$