# In a circuit, do surface charges get influenced by external electric fields? [duplicate]

Hello ,This image represents surface charges in a circuit caused by the battery's electric field, if you were to place an other electric field source of the same magnitude (but not connecting it to the circuit (on top of the battery or under it)) and placing so it has the same direction of the battery's electric field for me it would do 2 things :

First, since they are in the same direction their power will add up resulting in a doubled potential difference (e.g voltage) (well not totally since the distance at least a bit more), this will translate to the surfaces charges as a double of their density everywhere to maintain the same electric field inside the wire.

Second, The battery will get used more quickly due to the higher current.

However I know that this is wrong because :

This requires if I place the other electric field source to counter act the electric field of the battery, or if the other electric field is stronger will imply many weird reactions(after thinking about it you need a really strong second battery) .

The evenements I describe don't seem to really happen in the real life,

So if someone could show me the error in my reasoning,Thanks. TLDR: In a circuit why dosen't the electric field of other external things even help a tiny bit when stacking with the battery's electric field.

## marked as duplicate by John Rennie electromagnetism StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 23 at 5:08

• It actually did awnser my question – mohamed azaiez Jul 22 at 9:49

• The answer you linked to is not inconsistent with mine. Yes, if you bring some charge near one of the plates of a capacitor, it will induce charges and change the potentials on the plates. However the fact remains that any surface charge has the two components I mentioned earlier, which can be superposed. The surface charge due to external fields won't impact the circuit operation. The I-V relation of a capacitor is $I = C\frac{dV}{dt}$. Since the voltage difference due to the external field is not time-varying (or is slowly-varying) the circuit will operate as if there were no external field. – Puk Jul 8 at 6:16