Question

An oil drop carrying a charge of 2 electrons has a mass 3.2 into 10 raised minus 17 kg it is falling freely in air with terminal speed. The electric field

Answer 1

Q) An oil drop carrying a charge of 2 electrons has a mass of 3.2×$10^{-17}$ kg. It is falling freely into the air with terminal speed. The electric field required to make the drop move upwards with the same speed is (g=10$ms^{2}$ ).

(A)2×$10^{3}$V$m^{-1}$

(B)4×$10^{3}$V$m^{-1}$

(C)3×$10^{3}$V$m^{-1}$

(D)8×$10^{3}$V$m^{-1}$

Option (A) 2 × $10^{3}$ V$m^{-1}$ is the answer.

Given,

The charge of oil = 2electrons.

Mass = 3.2 * $10^{-17}$.

To Find,

The electric field is required to make the oil drop move upwards.

Solution,

Given that,

The oil drop has a mass of 3.2 * $10^{-17}$ kg.

When the forces acting on the drops balances the terminal velocity will attain.

So,

2mg = qE

m = 3.2 * $10^{-17}$

g = 10m$s^{2}$

q = 2

E =?

E = $\frac{2* 3.2 * 10^{-17}k * 10 }{2*(1.6 * 10^{-19} }$

E = 2 * $10^{3}$ V$m^{-1}$.

Hence, 2 × $10^{-3}$V$m^{-1}$ is the electric field needed for the oil drop to move upward.

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Answer 2

Answer:

The electric field is given by: $2 * 103 V/m$

Explanation:

We know,

Terminal velocity is achieved when Forces balance each other.

Here,

$m g=q E$

$3.2 \times 10^{-17} \times \mathrm{g}=4 \times 1.6 \times 10^{-19} \times \mathrm{E}$

$\mathrm{E}=2 \times 10^{3}$

Now,

Whether it be upwards velocity or downward velocity the condition of terminal velocity is same.

Hence, the Electric field will be same for both the cases. 2 × 103 V/m