Electromagnetism – Does Electric Current Flow from Higher Potential to Lower Potential?

electric-fieldselectromagnetismelectrostaticspotentialpotential energy

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Here, $O$ is a field charge with a $+T$ charge. $A$ and $B$ are test charges with $+T_1$ and $+T_2$ charge, and they are fixed in place. Suppose the points at which $A$ and $B$ are situated are $P_1$ and $P_2$ respectively. Let $T_2>T_1$ and $r_2>r_1$ such that the electric potential energy of $B$ becomes greater than that of $A$. However, clearly, the electric potential of $P_1$ is greater than that of $P_2$ as $r_1<r_2$.

If I connect $A$ and $B$ using a wire, will electric current flow from $A$ to $B$ or from $B$ to $A$?

My hypothesis:

I hypothesize that current will flow from $B$ to $A$ until the potential energies of $A$ and $B$ become equal.

My question:

  1. Is my hypothesis correct?

Best Answer

Electric fields exert forces on charges. The charges then respond to these forces according to Newton’s laws.

Suppose you have two objects with charges $q_1\neq q_2$ and masses $m_1\neq m_2$, initially at rest. Coulomb’s Law tells you the magnitude and direction of the force on each charge. In order to conserve momentum, the less-massive charge will do more of the moving. This is why, for instance, we can approximate the nucleus as “stationary” when we discuss atoms.

If you connect two charges with a macroscopic conductor, you can generally consider the motion of the charge carriers without worrying about the back-reaction of neutral bulk of the conductor — just like, when you blow out a candle, you don’t generally have to treat your body like a rocket engine which recoils away from the candle.

In metallic conductors the current is carried by electrons which flow towards high potentials — that is, antiparallel to the electric field. But in electrolytes, it can be that the positive charges have more mobility than the negative charges.

If I understand your specific scenario, you have (collapsing to one dimension)

(Q)              (q)         (2q)
 O                A           B

where the charge $Q$ at the origin is much larger the charges $q,2q$ at $A$ and $B$. You weren’t specific about the ratio of charges, here $q_B = 2q_A$, so I picked one arbitarily; you also weren’t specific about the ratio of distances, except that $r_B > r_A$.

If you connect your two “test” charges with a conductor, you’ll have two charged objects instead of three:

(Q)              (      3q     )
 O                A———————————B

The charges on the surface of the wire will distribute themselves so that the electric field on the inside of the wire is zero. Exactly how this is done can be quite complicated, especially if the wire isn’t straight: electric charges have a habit of accumulating on the high-curvature parts of surfaces, like corners or like the pointy ends of lightning rods. In general the combined positively-charged object $A—B$ will be more negative at its $A$ end and more positive at its $B$ end.

You set up $B$ to be more positive than $A$ in your initial condition; therefore the question of whether charge flows from $A$ to $B$ or vice-versa is quantitative, rather than qualitative.