# Calculation of radius

1. Nov 10, 2015

### gracy

1. The problem statement, all variables and given/known data
A spherical drop of water carrying a charge of 3×10^-19 has a potential of 500V at its surface (with V=0 at infinity) (a) What is the radius of the drop? If two such drops of the same charge and radius combined to form a single drop, what is the potential at the surface of the new drop

2. Relevant equations
$V$=$\frac{q}{4πε0r}$

3. The attempt at a solution
Here q=3×10^-19 C and V=500 volts.
$r$=$\frac{3×10^-19×9×10^9}{500}$
=$\frac{27×10^-10}{5×10^2}$
=$\frac{27×10^-10×10^-2}{5}$
=$\frac{27×10^-12}{5}$
=5.4×10^-12
but the answer for r is wrong it should be r=0.54cm.What went wrong .

2. Nov 10, 2015

### BvU

Typo (or hasty reading ) : 3 x 10 -19 should have been 3 x 10 -9.

For comparison: 1 single electron has a charge -1.60217662 × 10-19 Coulomb

3. Nov 10, 2015

### gracy

Neither of them.Actually it is
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:
:
:
:
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Printing mistake in the textbook.

4. Nov 10, 2015

### BvU

That's what we call a typo . The printer prints what has been typeset

5. Nov 10, 2015

### gracy

I just wanted to make sure that's not my mistake

6. Nov 10, 2015

### BvU

Part (b) is going OK ?

7. Nov 10, 2015

### gracy

Yes!
seriously, loved the way you paid attention to my all the threads today.