Electric Potential A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

Objectives: After completing this module, you should be able to: Understand an apply the concepts of electric potential energy, electric potential, and electric potential difference. Calculate the work required to move a known charge from one point to another in an electric field created by point charges. Write and apply relationships between the electric field, potential difference, and plate separation for parallel plates of equal and opposite charge.

Review: Work and Energy Work is defined as the product of displacement d and a parallel applied force F. Work = Fd; Units: 1 J = 1 N m Potential Energy U is defined as the ability to do work by virtue of position or condition. (Joules) Kinetic Energy K is defined as the ability to do work by virtue of motion (velocity). (Also in joules)

Signs for Work and Energy Work (Fd) is positive if an applied force F is in the same direction as the displacement d. The force F does positive work. A B The force mg does negative work. m F d The P.E. at B relative to A is positive because the field can do positive work if m is released. mg P.E. at A relative to B is negative; outside force needed to move m.

Gravitational Work and Energy Consider work against g to move m from A to B, a vertical height h. A B m F h Work = Fh = mgh mg g At level B, the potential energy U is: U = mgh (gravitational) The external force does positive work; the gravity g does negative work. The external force F against the g-field increases the potential energy. If released the field gives work back.

Electrical Work and Energy An external force F moves +q from A to B against the field force qE. B + + + + - - - - A d Fe Work = Fd = (qE)d + +q At level B, the potential energy U is: qE E U = qEd (Electrical) The E-field does negative work; External force does positive work. The external force F against the E-field increases the potential energy. If released the field gives work back.

Work and Negative Charges Suppose a negative charge –q is moved against E from A to B. B + + + + - - - - A d Work by E = qEd qE -q At A, the potential energy U is: E U = qEd (Electrical) No external force is required ! The E-field does positive work on –q decreasing the potential energy. If released from B nothing happens.

Work to move +q from A to B. Work to Move a Charge Work to move +q from A to B. + Q ¥ qE F rb · A B At A: ra At B: Avg. Force: Distance: ra - rb

Absolute Potential Energy Absolute P.E. is relative to ¥. + Q ¥ qE F · A B ra rb It is work to bring +q from infinity to point near Q—i.e., from ¥ to rb Absolute Potential Energy:

Positive potential energy Example 1. What is the potential energy if a +2 nC charge moves from ¥ to point A, 8 cm away from a +6 mC charge? The P.E. will be positive at point A, because the field can do + work if q is released. +6 mC +Q · A +2 nC 8 cm Potential Energy: U = 1.35 mJ Positive potential energy

Signs for Potential Energy Consider Points A, B, and C. +6 mC +Q · A 8 cm · B C 12 cm 4 cm For +2 nC at A: U = +1.35 mJ If +2 nC moves from A to B, does field E do + or – work? Does P.E. increase or decrease? Questions: +2 nC Moving positive q The field E does positive work, the P.E. decreases. If +2 nC moves from A to C (closer to +Q), the field E does negative work and P.E. increases.

Note that P.E. has decreased as work is done by E. Example 2. What is the change in potential energy if a +2 nC charge moves from A to B? +6 mC +Q · A 8 cm B 12 cm Potential Energy: From Ex-1: UA = + 1.35 mJ DU = -0.450 mJ DU = UB – UA = 0.9 mJ – 1.35 mJ Note that P.E. has decreased as work is done by E.

Moving a Negative Charge Consider Points A, B, and C. +6 mC +Q · A 8 cm B C 12 cm 4 cm Suppose a negative -q is moved. If -q moves from A to B, does field E do + or – work? Does P.E. increase or decrease? Questions: Moving negative q - The field E does negative work, the P.E. increases. What happens if we move a –2 nC charge from A to B instead of a +2 nC charge. This example follows . . .

(Negative due to – charge) Example 3. What is the change in potential energy if a -2 nC charge moves from A to B? Potential Energy: +6 mC +Q · A 8 cm B 12 cm From Ex-1: UA = -1.35 mJ (Negative due to – charge) UB – UA = -0.9 mJ – (-1.35 mJ) DU = +0.450 mJ A – charge moved away from a + charge gains P.E.

The field E exist independently of the charge q and is found from: Properties of Space An electric field is a property of space allowing prediction of the force on a charge at that point. E Electric Field + Q . r The field E exist independently of the charge q and is found from: E is a Vector

Electric Potential Electric Potential: Electric potential is another property of space allowing us to predict the P.E. of any charge q at a point. Potential + Q . r P Electric Potential: The units are: joules per coulomb (J/C) For example, if the potential is 400 J/C at point P, a –2 nC charge at that point would have P.E. : U = qV = (-2 x 10-9C)(400 J/C); U = -800 nJ

The SI Unit of Potential (Volt) From the definition of electric potential as P.E. per unit charge, we see that the unit must be J/C. We redefine this unit as the volt (V). A potential of one volt at a given point means that a charge of one coulomb placed at that point will experience a potential energy of one joule.

Calculating Electric Potential Electric Potential Energy and Potential: Potential + Q . r P Substituting for U, we find V: The potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.)

Since P.E. is positive, E will do + work if q is released. Example 4: Find the potential at a distance of 6 cm from a –5 nC charge. Q = -5 nC - Q . r P 6 cm q = –4 mC Negative V at Point P : VP = -750 V What would be the P.E. of a –4 mC charge placed at this point P? U = qV = (-4 x 10-6 mC)(-750 V); U = 3.00 mJ Since P.E. is positive, E will do + work if q is released.

Potential For Multiple Charges The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge. + - · Q1 Q2 Q3 A r1 r3 r2 Potential is + or – based on sign of the charges Q.

Example 5: Two charges Q1= +3 nC and Q2 = -5 nC are separated by 8 cm Example 5: Two charges Q1= +3 nC and Q2 = -5 nC are separated by 8 cm. Calculate the electric potential at point A. + · Q2 = -5 nC - Q1 +3 nC 6 cm 2 cm A B VA = 450 V – 2250 V; VA = -1800 V

Example 5 (Cont.): Calculate the electric potential at point B for same charges. + · Q2 = -5 nC - Q1 +3 nC 6 cm 2 cm A B VB = 1350 V – 450 V; VB = +900 V

Example 5 (Cont.): Discuss meaning of the potentials just found for points A and B. Consider Point A: VA = -1800 V + · Q2 = -5 nC - Q1 +3 nC 6 cm 2 cm A B For every coulomb of positive charge placed at point A, the potential energy will be –1800 J. (Negative P.E.) The field holds on to this positive charge. An external force must do +1800 J of work to remove each coulomb of + charge to infinity.

Example 5 (Cont.): Discuss meaning of the potentials just found for points A and B. + · Q2 = -5 nC - Q1 +3 nC 6 cm 2 cm A B Consider Point B: VB = +900 V For every coulomb of positive charge placed at point B, the potential energy will be +900 J. (Positive P.E.) For every coulomb of positive charge, the field E will do 900 J of positive work in removing it to infinity.

Potential Difference The potential difference between two points A and B is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential. Potential Difference: VAB = VA - VB WorkAB = q(VA – VB) Work BY E-field The positive and negative signs of the charges may be used mathematically to give appropriate signs.

Example 6: What is the potential difference between points A and B Example 6: What is the potential difference between points A and B. What work is done by the E-field if a +2 mC charge is moved from A to B? + · -5 nC - Q1 +3 nC 6 cm 2 cm A B Q2 VA = -1800 V VB = +900 V VAB= VA – VB = -1800 V – 900 V Note point B is at higher potential. VAB = -2700 V WorkAB = q(VA – VB) = (2 x 10-6 C )(-2700 V) Work = -5.40 mJ E-field does negative work. Thus, an external force was required to move the charge.

Example 6 (Cont.): Now suppose the +2 mC charge is moved from back from B to A? · -5 nC - Q1 +3 nC 6 cm 2 cm A B Q2 VA = -1800 V VB = +900 V VBA= VB – VA = 900 V – (-1800 V) This path is from high to low potential. VBA = +2700 V WorkBA = q(VB – VA) = (2 x 10-6 C )(+2700 V) Work = +5.40 mJ E-field does positive work. The work is done BY the E-field this time !

Parallel Plates Consider Two parallel plates of equal and opposite charge, a distance d apart. VA + + + + - - - - VB E +q F = qE Constant E field: F = qE Work = Fd = (qE)d Also, Work = q(VA – VB) So that: qVAB = qEd and VAB = Ed The potential difference between two oppositely charged parallel plates is the product of E and d.

Example 7: The potential difference between two parallel plates is 800 V. If their separ- ation is 3 mm, what is the field E? VA + + + + - - - - VB E +q F = qE The E-field expressed in volts per meter (V/m) is known as the potential gradient and is equivalent to the N/C. The volt per meter is the better unit for current electricity, the N/C is better electrostatics.

Summary of Formulas Electric Potential Energy and Potential Electric Potential Near Multiple charges: WorkAB = q(VA – VB) Work BY E-field Oppositely Charged Parallel Plates:

CONCLUSION: Chapter 25 Electric Potential