Chapter 17 Electrical Energy and Current 17 – 1 Electric Potential

Review of gravitational potential energy As a mass moves from point A to point B, the gravitational field does work on a mass: WAB=F d = mgd The change in gravitational potential energy is equal to the work done by the gravity: ΔPE = W AB= mgΔh

Electric potential energy Electrical potential energy can be associated with a charge in a uniform field. ΔPEelectric = qEd magnitude of change in PE q = charge E = electric field d = displacement from a reference point

ME = KE + PEgrav + PEelastic + PEelectric Electrical potential energy is a component of mechanical energy. ME = KE + PEgrav + PEelastic + PEelectric

Consider the case of a charge moving in an electric field As the positive charge moves from A to B, work is done WAB = F d = q E d ΔPE = W AB= q E Δd Moving a positive test charge against the direction of an electric field is like moving a mass upward within Earth's gravitational field. Both movements would be like going against nature and would require work by an external force. This work would in turn increase the potential energy of the object. On the other hand, the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy.

Energy and Charge Movements Direction of movement + charge - charge Along E Loses PE Gains PE Opposite E Moving a positive test charge against the direction of an electric field is like moving a mass upward within Earth's gravitational field. Both movements would be like going against nature and would require work by an external force. This work would in turn increase the potential energy of the object. On the other hand, the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy.

Electric potential energy for a pair of point charges If the charges have the same sign, PE is positive. The force would be repelling. Positive work must be done to force the two charges near one another. If the charges have opposite signs, PE is negative. The force would be attractive. Work must be done to hold back the unlike charges (left on their own they would accelerate as they come closer together. Note: The reference point for zero potential energy is usually at 

Example What is the potential energy of the charge configuration shown? q 3 = 1.0 mc 0.30 m 0.30 m q 2 = - 2.0 mc q 1 = 2.0 mc 0.30 m

Given: q 1 = 2.0 x 10-6 C q 2 = - 2.0 x 10-6 C q 3 = 1.0 x 10-6 C r = 0.30 m (same for all) k= 9.0 x 109 n•m2/C2 The total potential energy is the algebraic sum of the mutual potential energies of all pairs of charges.

PET = PE12 + PE23 + PE 13 = k q1q2 + kq2q3 + kq1q3 r r r = k [(q1q2) + (q2q3 ) + (q1q3)] r = [(9.0 x 109 n•m2/C2)/ 0.30] x [(2.0 x 10-6 C)(- 2.0 x 10-6 C) + (- 2.0 x 10-6 C)(1.0 x 10-6 C) + (2.0 x 10-6 C)(1.0 x 10-6 C)] =- 0.12 J

Electric Potential Since the electrical potential energy can change depending on the amount of charge you are moving, it is helpful to describe the electrical potential energy per unit of charge. 

Electric Potential  

 

If a charge moves between two points in an electric field, then it undergoes a change in potential, or potential difference given by

Potential Difference, continued The potential difference in a uniform field varies with the displacement from a reference point. ∆V = - Ed Potential difference in a uniform electric field E = electric field d = displacement in the field

Sample Problem A proton moves from rest in an electric field of 8.0104 V/m along the +x axis for 50 cm. Find a) the change in in the electric potential, b) the change in the electrical potential energy, and c) the speed after it has moved 50 cm.

a) V = -Ed = -(8.0104 V/m)(0.50 m) = -4.0104 V b) PE = q V = (1.610-19 C)(-4.0 104 V) = -6.4 10-15 J C) Ei = Ef KEi+PEi = KEf + PEf, since KEi=0 KEf = PEi – PEf = -PE 1/2 mpv2 = -PE v = 2 PE/m = 2(6.4x10-15 J)/1.67x10-27 kg)=2.8x106 m/s

Potential Difference, point charges Chapter 17 Potential Difference, point charges At right, the electric potential energy at point A depends on the charge at point B and the distance r. An electric potential exists at some point in an electric field regardless of whether there is a charge at that point.

The reference point for potential difference near a point charge is often at infinity. The potential difference between a point at infinity and a point near a point charge is given by:

Electric Potential of Multiple Point Charges- Superposition The total electric potential at point “a” is the algebraic sum of the electric potentials due to the individual charges.

Problem Solving with Electric Potential (Point Charges) Remember that potential is a scalar quantity So no components to worry about  Use the superposition principle when you have multiple charges Take the algebraic sum Keep track of sign The potential is positive if the charge is positive and negative if the charge is negative Use the basic equation V = kcq/r

Example: Finding the Electric Potential at Point P = -2.0 mC 5.0 mC =

Superposition: Vp=V1+V2