Question

3. In an oil drop experiment, a drop with a weight of 1.9x10-14 N was suspended motionless when the potential difference between the plates that were 63 mm apart was 0.78 kV. What was the charge on the drop? (Positive plate on the top)

Answer 1

Answer:

The charge on the drop is 1.50$\times$ 10^-6.

Explanation:

Step 1: Given that, particle is at rest and so net force is zero.

From the free body diagram we can see that

$\begin{aligned}& F_e=F_g \\& \mathrm{mg}=\mathrm{q} \frac{\mathrm{V}}{\mathrm{d}}=(\mathrm{ne}) \frac{\mathrm{V}}{\mathrm{d}}\end{aligned}$

where q=n e ( n is Number of electrons)

Step 2:Substituting the given values of-

m= 1.9x10^-14 N

V =0.78x10^3 N

d=0.063 m

e=  1.602×10 ^−19 Coulomb

$\frac{\left(1.9 \times 10^{-14}\right)(9.8)\left(63 \times 10^{-3}\right)}{\left(1.602 \times 10^{-19}\right)\left(0.78 \times 10^3\right)}$

= 1.50$\times$ 10^-6

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