Electric Potential & Potential Energy

Transcription

1 Electric Potential & Potential Energy I) ELECTRIC POTENTIAL ENERGY of a POINT CHARGE Okay, remember from your Mechanics: Potential Energy (U) is gaine when you o work against a fiel (like lifting a weight, for example). Work is (+) whenever U is lost an K is gaine (remember, W = K). As a result, U is gaine whenever work is (-) (i.e., whenever, K is lost). More remembering: (a) The expression for electrostatic force, aka Coulomb's Law, is given by F el = kqq'/r 2 (b) Since this is a force that varies with position, the work one by this force in moving a point charge q from to a istance r away from a charge q is given by (compliments of the Work/Energy & Universal Gravitation Notes): W = kqq'/x 2 x U el = -kqq' x/x 2 r U el = kqq'/r Base on the expression above for U el, we can efine the electric potential energy (Uel) relative to infinity between two point charges q an q' that are separate by a istance r as U el = kqq'/r For the charge configuration shown below, fin the total electric potential energy. Assume q = 1.0 x 10-7 C an a = 10 cm. -4q a a +q a +2q

2 II) ELECTRIC POTENTIAL (V) at a point, POTENTIAL DIFFERENCE, & WORK Electric Potential Due to a Point Charge Electric potential represents the number of Joules of electric potential energy (U el ) per Coulomb of point charge q that is in an electric fiel cause by a charge q'. Therefore, by using the expression for U el, we get V = kq'/r (where q' causes an electric fiel at a istance r away) The unit for potential, V, is the Volt: 1 Volt = 1Joule/1 Coul There are 2 qualitative relationships between potential V an electric fiel, E: (1) The value for V ecreases if you move in the irection of the electric fiel lines. + - Here, V becomes less (+) as you move out along the fiel lines. Becoming less (+) is a ecrease. Here, V becomes more (-) as you move in along the fiel lines. Becoming more (-) is a ecrease. (2) The lines of equipotential are always perpenicular to the electric fiel lines + - The ashe lines are lines of equipotential. At any point along these lines, the value for V is the same. An, the electric fiel, as you can see, is always to the equipotential lines. This is true at any point, regarless of the shape of the fiel.

3 Potential Difference (Voltage) A given value for V, by itself, oesn t tell us much it s more like a reference point. We nee it to calculate something that s of greater use the potential ifference. The potential ifference ( V), or voltage, in moving charge q from point A to point B while it is uner the influence of charge(s) q' is foun by V = V AB = V B - V A (Remember, the unit is Volts) Change in Potential Energy: The External Work Done in Moving a Charge Potential ifference can be a slippery slope since you can have a loss of potential for a charge that gains potential energy, an a gain of potential for a charge that loses potential energy. This is where the external work one in moving a charge comes in. The external work one in moving a charge is like the work one by an applie force in lifting or lowering a weight: it represents the change in its potential energy. It s (-) when it s move the way it wants to go naturally, an (+) when it s move the way it oesn t want to go naturally. It involves both the sign of the charge being move an the sign of the potential ifference. When we combine these two factors, we get a result (for external work) that makes sense. The external work one in moving a charge q from point A to point B is foun by: See if these situations makes sense : W = q( V) = U + V is (-), W is (-), U is (-) & K is (+) + E - V is (-), W is (+), U is (+) & K is (-) + - E - V is (+), W is (+), U is (+) & K is (-) V is (+), W is (-), U is (-) & K is (+) What about if you re aske to fin the work one by the electric fiel? It s just the opposite of the external work. So, if the external work ( U) is -5 J, then the work one by the fiel is + 5 J.

4 2.2 - For the charge istribution shown below, fin: (a) the work one in moving the +2 C charge from to the position below (b) the work one by the fiel +2 C 10 cm 10 cm +1 C +1 C 10 cm For the charge istribution shown, Fin (a) the potential at point A (b) the work one in moving a charge of q = 1 C from point A to point B, locate miway between the charges. A 3 cm 4 cm +4 C -6 C B

5 III) ELECTRIC KINETIC ENERGY When a charge is release in an electric fiel, it unergoes a change in its potential an therefore a change in its potential energy (remember the gravity analog... it's like ropping a ball or tossing it up in the air; there's a change in U g ). This change in U el "turns into" a change in K (remember the Conservation of Energy a loss of U el becomes an equal gain in K). Therefore, use the Conservation of Energy to fin the spee of a charge particle that unergoes a change in its electric potential energy. Since Then U el = qv (from Section II) U el = q V K = q V or (1/2)mv f 2 (1/2)mv 0 2 = q V In using this relationship, remember that if q V is (+), then K will be (-); if q V is (-), then K will be (+). A gain in potential energy correspons to a loss of kinetic energy, an vice-versa. You make the sign ajustment. If you on t like worrying about signs, you can use conservation of energy In particular U0 + K0 = Uf + Kf qv0 + K0 = qvf + Kf revisite Suppose the + 2 C is release from rest. (a) Describe its subsequent motion (the AP folks like to ask this ) (b) Fin its spee as r (i.e., when it s really far away).

6 IV) CONNECTING the FIELD E to the POTENTIAL V So far, we ve ealt only with point charges. But if the charges are istribute over a surface or shape, then we must integrate to fin V or V. Since U = - F(x) x, then U/q = V = - (F/q) x = - E x. Possibility 1: The electric fiel is uniform (this inclues the sheet/plate of charge case) V = - E x = Ex V = E A positive charge +q moves (as shown) through a isplacement from point A to point B in a uniform electric fiel E. Fin its potential ifference an the work one by the fiel. E A +q B These are equipotential lines. The value for V is the same at any point along them, AND they re perpenicular to the electric fiel lines Repeat the previous problem, but consier a negative charge, -q. E A -q B Fin the potential ifference in moving a (+) test charge a istance "" from left to right between the oppositely charge parallel plates, as shown below. Both plates have a charge istribution. (Remember, from Gauss' Law, that fiel E = for a charge conucting plates) E +q + -