A water drop of mass 3-2times 10-18- kg and carrying a charge of 1-6times 10-19-C is suspended stationary between two plates of an electric field- Given g-10 m-s-2- the intensity of the electric field required is2 V-m200 V-m20 V-m2000 V-m

Question

Toppr Admin · Accepted Answer

Force in electric fieldF-qEand F-mg-xA0-xA0-xA0- -By gravitation-So-mg-qE -xA0- -for equilibrium-E-mgq-xA0- -xA0- -3-2-xD7-10-x2212-18-xD7-101-6-xD7-10-x2212-19-xA0- -xA0- -200V-m

A water drop of mass $3.2 \times 10^{- 18}$ kg and carrying a charge of $1.6 \times 10^{- 19}$ C is suspended stationary between two plates of an electric field. Given $g = 10$ $m / s^{2}$ , the intensity of the electric field required is
$2 V / m$
$200 V / m$
$20 V / m$
$2000 V / m$

Force in electric field
$F = q E$
and $F = m g$ (By gravitation)
$S o, m g = q E$ (for equilibrium)

$E = \frac{m g}{q}$

$= \frac{3.2 \times 10^{- 18} \times 10}{1.6 \times 10^{- 19}}$

$= 200 V / m$

A water drop of mass 3.2×10−18 kg and carrying a charge of 1.6×10−19C is suspended stationary between two plates of an electric field. Given g=10 m/s2, the intensity of the electric field required is2V/m200V/m20V/m2000V/m

Force in electric fieldF=qEand F=mg (By gravitation)So,mg=qE (for equilibrium)E=mgq =3.2×10−18×101.6×10−19 =200V/m

A water drop of mass $3.2 \times 10^{- 18}$ kg and carrying a charge of $1.6 \times 10^{- 19}$ C is suspended stationary between two plates of an electric field. Given $g = 10$ $m / s^{2}$ , the intensity of the electric field required is
$2 V / m$
$200 V / m$
$20 V / m$
$2000 V / m$

Force in electric field
$F = q E$
and $F = m g$ (By gravitation)
$S o, m g = q E$ (for equilibrium)

$E = \frac{m g}{q}$

$= \frac{3.2 \times 10^{- 18} \times 10}{1.6 \times 10^{- 19}}$

$= 200 V / m$