Why do electrons move from lower to higher electric potential?

[theonlytycrane]

In my diagram I have a positive source charge Q. I would like to compare the difference between how positive test charge A and negative test charge B respond to the Electric field created by Q. Let's ignore the effects of A and B on one another.

Positive test charge A moves away from Q, and as it does so it's electric potential V decreases.

Negative test charge B moves toward Q, and as it does so it's electric potential V increases. This is the point that I am confused on.

In both cases, the electric potential energy decreases, as A moves farther away from Q and B moves towards Q. Is the movement from lower to higher electric potential for the negative test charge just a result of the convention used in defining electric potential? Is the fact that potential energy is decreasing in both instances the important point to note?

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[RegalDan]

Break down the equation for electric potential and it will become rather clear. Electric potential: V = U/q, and even further, U = kQq/r. For an electron, the sign of k is positive, since it is a constant, the sign of Q is positive, since the test charge is positive, and the sign of q is negative, since the point charge is an electron the value is -1.6x10^-19 C. r is the distance between the test charge and the point charge, and is a positive value, since distance can not be negative. This means that overall, U is negative for an electron. Plug this information into the equation for electric potential: V = -U/q, and remember that q in this case is negative because of the negative charge that an electron carries, then you will get: V = -U/-q, which equals a positive value for V. I hope this helps a little.

Edit: To answer the question you named the post after, look at the value of r. An electron will move close to a positive test charge (Q), decreasing the value of r. A decreasing denominator will increase the value of V. Ex: if U = 10 (simple, for the sake of the example) and an electron moves from a distance of r = 5 to r = 2, V will change from 2 to 5 respectively, increasing the value of V as r decreases.

 

[theonlytycrane]

Thanks! This helps to make sure that I am thinking about it correctly.

@RegalDan- one note to make is that you meant to say the sign of Q is positive since the source charge is positive.

 

[RegalDan]

@RegalDan- one note to make is that you meant to say the sign of Q is positive since the source charge is positive.

Nice save, hope it didn't add too much confusion, but I think you've got it!

 

[azor ahai]

think of a circuit. current flows from high charge density (potential/voltage) to low charge density (potential/voltage). current uses (+) test charges because they're stupid, and so actual electron flow is reversed. So electrons move from low (+) charge density or voltage because there's a high (-) charge there, and moves to high (+) charge density because there's a low (-) charge there. Like charges repel.

 

[ElectricNoogie]

Electrical potential energy is analogous to gravitational potential energy. Charges, like masses, will want to go wherever their potential E is lowest, like a ball rolling down hill. You don't need to do anything, the ball will roll down on its own. For charges however, we must remember that, unlike masses, the forces can work in two directions. Like charges repel each other and that's what they will do to minimize EPE. Opposite charges will attract each other and move closer to minimize their EPE.

By convention, when the MCAT says test charge, they mean + charge.

When deciding what should happen to EPE and work, ask yourself "Is the charge going where it WANTS to go (to lower EPE)? If YES, you don't need to do any work (W < 0) and EPE will drop. If NO, then you need to put work into the charge to get it there (W >0) and EPE of the charge will rise.

Hop this helps, Good luck!