Question

125 identical drops each charged to the same potential of 50 volts are combined to form a single drop. the potential of new drop will be

Answer 1

Answer:

V1 = 4 pi / 3 * R1^3        assuming spherical drops

V2 = 125 V1 = 4 pi / 3 * R2^3

V2 / V1 = 125 = R2^3 / R1^3

R2 / R1 = 125^1/3 = 5

V1 = k Q1 / R1      electric potential of 1 drop

V2 = k Q2 / R2

V2 / V1 = (Q2 / Q1) * ( R1 / R2) = 125 / 5 = 25

V2 = 25 V1 = 25 * 50 = 1250 V

Answer 2

Answer:

The potential of the new drop will be 1250V.

Explanation:

Let the radius of the single drop be R and that of small identical drops be r. Then,

$\frac{4}{3}\pi\times R^3=125\times\frac{4}{3}\pi\times r^3$

$R=5r$              (1)

And the potential is given as,

$V=\frac{Q}{4\pi\epsilon r}$         (2)

Where,

V=potential

Q=charge on the sphere

r=radius of the sphere

ε=permitivity of the medium

The potential for small drops is given as,

$50=\frac{ Q}{4\pi\epsilon\times r}$             (3)            (V=50v given in question)

Potential on the single drop,

$V_2=\frac{q}{4\pi\epsilon\times R}=\frac{ 125\times Q}{4\pi\epsilon\times 5r}$        (4)           (using equation (1) and also q=125Q)

By using equation (3) in equation (4) we get;

$V_2=\frac{125}{5}\times 50=1250V$

Hence, the potential of the new drop will be 1250V.