# Re: [AMBER] Amber16 electric field parameters

From: Pengfei Li <ambermailpengfei.gmail.com>
Date: Tue, 3 Oct 2017 21:10:00 -0500

Hi Alessandro,

> On Oct 2, 2017, at 8:52 AM, Ross Walker <ross.rosswalker.co.uk> wrote:
>
>>
>> On Oct 2, 2017, at 4:18 AM, Alessandro Mariani <alessandro.mariani.uniroma1.it> wrote:
>>
>> Hello,
>> as far as I can get, the conversion is something like:
>> 1 kcal/(mol*A*e) = 4.184 kJ/(mol*A*e) =
>> (4184/1.6022*10**-19)*V/(mol*A) = (4184/1.6022*10**-29)*V/(mol*m)
>> = 2.61*10**32 V/(mol*m)
>>
>> Now i really do not understand why that "mol" is there. What is it supposed
>> to be? Avogadro's number? The number of moles in the box? The number of
>> molecules in the box? Anyways, a conversion factor in 10**32 is
>> astonishing, and even if mol is substituted with Avogadro's number, I
>> get 1 kcal/(mol*A*e)
>> = 4.34*10**8 V/m that is a nonsense to me. Can please someone give me an
>> hint? And, maybe, explain to me why to not use SI units in Amber?
>

I think your calculation has no problem.

1 (kcal/mol*A*e) = 4184.0 / 1.6022 / 6.022 * 10^6 V/m, which is about 4.34 * 10^8 V/m.

In fact, people like to use MV/cm unit under this kind of situation.

1 MV/cm = 1*10^8 V/m

So 1 kcal/(mol*A*e) is about 4.34 MV/cm.

Kind regards,
Pengfei

>
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Received on Tue Oct 03 2017 - 19:30:02 PDT
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