Question

64 drops of radius 0.02m and each carrying a charge 5 microcoloumb are combined to form a bigger drop. Find how surface density electrification will change if no charge lost
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Answer 1

Let us consider,

r = radius of small drop

R = radius of large drop

Then, as per the question, we can say;

$64$ × $\frac{4}{3}\pi r^3$ $=$  $\frac{4}{3}\pi R^3$

⇒ $64r^3 = R^3$

⇒ $4r = R$ ...(Eqn. 1)

Then,

$r_1 = \frac{q}{A} = \frac{64 * 5 * 10^-^6}{64\pi r^2} = \frac{3125}{\pi } * 10^-^6$

And,  $r_2 = \frac{q}{A} = \frac{64 * 5 * 10^-^6}{4\pi R^2}$

On substituting the value of R with Eqn. 1, we get;

$r_2 = = \frac{64 * 5 * 10^-^6}{4\pi 16r^2} = r_1 * 4 = \frac{12500}{\pi } * 10^-^6$

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Answer 2

Given : 64 drops of radius 0.02m and each carrying a charge 5 microcoloumb are combined to form a bigger drop.

To find : Find how surface density electrification will change if no charge lost.

solution : let r is the radius of smaller drop and R is the radius of bigger drop.

as it is given that 64 drops coalesce to form a bigger drop.

so, volume of 64 drops = volume of a bigger drop

⇒64 ×4/3 πr³ = 4/3 πR³

⇒64r³ = R³

⇒4r = R

⇒4 × 0.02 = R

⇒R = 0.08 m

net charge on the bigger drop, Q = 64q = 64 × 5 = 320μC

now surface charge density of smaller drop ,σ₁ = q/4πr²

= 5μC/{4π(0.02)²} ........(1)

and surface charge density of bigger drop

, σ₂ = Q/4πR²

= (320μC)/{4π(0.08)²} .....(2)

from equations (1) and (2) we get,

σ₁/σ₂ = 5μC/{4π(0.02)²}/(320μC)/{4π(0.08)²}

= (1/64) × (0.08)²/(0.02)²

= 1/64 × 16/1

= 1/4

Therefore surface charge density of bigger drop is four times of a smaller drop.

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