# Why is there only a drop in potential energy when charges flow through a resistor?

First, I read that if you move a charge (say a positive charge) against an electric field, its electric potential energy increases because you're doing work to move the charge to that position. I also read that electric potential is the amount of potential energy per charge. To specify my question, when a positive charge leaves the positive terminal of a battery, ( I know the charge is actually negative, but let's assume it is positive) shouldn't its potential energy already be decreasing with each increase in distance you move the charge away from the terminal? Because it's like moving a charge in the direction of its field, so potential energy should decrease, right? So shouldn't there already be a drop in voltage ( potential difference) before the charges even reach a resistor?

• Are you familiar with capacitors? I think it is more obviously explained using these. – Quantum spaghettification Jun 15 '16 at 9:26
• Not really, sadly. I'm trying to understand why there needs to be resistance at all for there to be electric potential difference – Joel Khoo Jun 16 '16 at 5:16

So shouldn't there already be a drop in voltage ( potential difference) before the charges even reach a resistor?

Normally we view a battery or a cell as accumulator of charges in a manner that a potential difference is built up when we charge a battery with plates and electrolyte. Therefore if charges flow out or current is being drawn at certain voltage one expects depletion in the reservoir but that variation in the voltage is dependent on the amount of energy drawn-if one unit of charge is moved out the variation will not be measurable.

I read that if you move a charge (say a positive charge) against an electric field, its electric potential energy increases because you're doing work to move the charge to that position. I also read that electric potential is the amount of potential energy per charge.

Now let us take up the view on "Potential Energy" of the electric field- One should try to visualize the space around a point charge/charge distribution and the points in coordinate space having a 'field intensity' which can be measured by a test charge and defined by intensity vector , based on coulomb's law. The vector field can also be mapped by a "scalar potential" which is attributed by the amount of work done in moving a test charge from the point in space where the field intensity is absent/zero (naturally at infinite distance) to the point where field potential is being calculated/defined.

Therefore one should try to understand the 'potential ' as a characteristic of the field rather than the particle.The charge field do provide EMF in driving the charges along a circuit in 'current electricity' and the difference of potential is a measure of the current times the resistance in the circuit- if no load is provided then the avalanche of charge flow may occur- and its a short circuit condition -discharging the battery in a short time interval through sparking/generation of heat etc.-which must be avoided.

• Does that mean that the charge does lose energy as it moves from positive to negative, but the difference is so small that it is negligible? – Joel Khoo Jun 20 '16 at 5:31
• Charge as it moves in a conducting wire is not like a stone falling from a height and in the process getting another form of mechanical energy.... as charges move it has to encounter many hurdles on the way...and the charge which started at one end might not reach the other end...at any temperature its motion gets a zig-zag path due to collision and one can only talk of drift velocity of the charges and a net transfer of charge at any vertical section...though the flow is driven by the potential drop. – drvrm Jun 20 '16 at 14:14
• Okay, I'm with you so far. But is the falling object analogy applicable for positive charges moved from infinity to a positively charged plate? As in, potential energy is gained by moving it closer to the plate and is lost by moving it away from the plate, right? Oh,and was my first comment right? That the difference is negligible or was I completely off? – Joel Khoo Jun 21 '16 at 0:54
• <potential energy is gained by( the particle) moving it closer to the plate and is lost by(the particle) moving it away from the plate, right?>... @Joel, I have a picture that no"internal changes" will be observed in both the situations as far as the particle is concerned,,, whereas the field /or agency which has done some work leads to 'defining' a potential in the space and converted to K.E. of the particle to do work....as the charges are at different potentials due to its environment it does move and share its energy.. in an ideal conductor the loss will be minimal. – drvrm Jun 21 '16 at 6:50

I think that you have missed an assumption that is often used in circuit theory and that is that the connecting leads have a negligible resistance.

In the real world the leads do have a resistance and so as the (positive) charge moves from the positive terminal of the battery there is a conversion of electric potential energy into heat and the same conversion happens as the charge passes through the resistor and the same again happens as the charge passes through the leads to the negative terminal of the battery.

However if the resistance of the leads is much less than the resistance of the resistor the conversion in the leads is very much smaller than in the resistor and so can be neglected.
From this it is assumed that there is no (really negligible) potential difference across the leads.

• I understand what you're saying. My teacher also explained it this way, but I'm more confused about the part where "moving a charge in its natural direction results in a loss in electric potential energy" and is like an object falling in the air. Shouldn't the charge, no matter if there is somehow no resistance in the circuit, lose potential energy as it moves from positive to negative? – Joel Khoo Jun 15 '16 at 23:27
• What happens to the potential energy of a marble rolling on a horizontal table? – Farcher Jun 18 '16 at 14:19
• I think I'm more confused with the explanation given here: physicsclassroom.com/Class/circuits/u9l1b.cfm#circuits – Joel Khoo Jun 21 '16 at 2:09
• <In a certain sense, an electric circuit is nothing more than an energy conversion system. In the electrochemical cells of a battery-powered electric circuit, the chemical energy is used to do work on a positive test charge to move it from the low potential terminal to the high potential terminal. Chemical energy is transformed into electric potential energy within the internal circuit (i.e., the battery).>.........@Joel , a quote from the above reference given by you ....pl. raise the issue of confusion. – drvrm Jun 21 '16 at 6:56
• "Similar reasoning would lead one to conclude that the movement of positive charge through the wires from the positive terminal to the negative terminal would occur naturally. Such a movement of a positive test charge would be in the direction of the electric field and would not require work. The charge would lose potential energy as moves through the external circuit from the positive terminal to the negative terminal." This is where it gets confusing for me. If the charge is continuously losing energy, shouldn't the voltage also drop continuously, regardless of any resistance? – Joel Khoo Jun 22 '16 at 0:08

Fist, the amplitude of electric field (thus the electric potential energy) in the wire is not uniform. The amplitude depends on the impedance of the place (I don't know the reason). For example, the eletric field is much stronger in resistor, capacitor or inductor.

Second, if the electric filed is weak, the drop in the potential will also be small. Since the the impedance of wire is quite small, the drop of potential is negligible.