Question

Two drops of mercury each has a charge on 0.0000000018 nC and a voltage of 434.21 V. If the two drops are merged into one drop, what is the voltage on this drop?

Answer 1

The other person is correct ;v

Answer 2

`V = 689.4 Volts`

Step-by-step explanation:

charge on the drop is given as

`Q = 0.0000000018 C`

V = 434.21 Volts

now we know that

`V = (kQ)/(r)`

so we have

`434.21 = ((9 * 10^9)(1.8 * 10^(-9)))/(r)`

`r = 3.73 cm`

now when two such drops is merged and forms a single drop then in that case

`V = V_1 + V_2`

`(4)/(3)\pi R^3 = 2((4)/(3)\pi r^3)`

so we have

`R = 2^(1/3) r`

`R = 4.7 cm`

now after merging two drops we have total charge on both drops is

`Q = 2q = 1.8 nC + 1.8 nC = 3.6 nC`

now potential of two drops is given as

`V = (KQ)/(R)`

`V = ((9 * 10^9)(3.6 * 10^(-9)))/(0.047)`

`V = 689.4 Volts`