Electric Potential A Power Point Presentation by Paul

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. B F m mg A The force F does positive work. The force mg does negative work d The P. E. at B relative to A is positive because the field can do positive work if m is released. 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. Work = Fh = mgh At level B, the potential energy U is: U = mgh (gravitational) B F m mg A h g 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 q. E. Work = Fd = (q. E)d At level B, the potential energy U is: U = q. Ed (Electrical) The E-field does negative work; External force does positive work. B ++++ Fe +q + d q. E E A - - 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. Work by E = q. Ed At A, the potential energy U is: U = q. Ed (Electrical) No external force is required ! B ++++ q. E -q d E A - - The E-field does positive work on –q decreasing the potential energy. If released from B nothing happens.

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

Absolute Potential Energy ++ + + ++Q++ rb Absolute P. E. is relative to ¥. B · A · ra Absolute Potential Energy: ¥ F+ q. E It is work to bring +q from infinity to point near Q—i. e. , from ¥ to rb 0

Example 1. What is the potential energy if a +2 n. C charge moves from ¥ to point A, 8 cm away from a +6 m. C charge? The P. E. will be positive at point A, because the field can do + work if q is released. Potential Energy: U = 1. 35 m. J A · 8 cm +2 n. C +Q +6 m. C Positive potential energy

Signs for Potential Energy Consider Points A, B, and C. For +2 n. C at A: U = +1. 35 m. J Questions: If +2 n. C moves from A to B, does field E do + or – work? Does P. E. increase or decrease? A · ·B 12 cm 8 cm +Q ·C 4 cm +6 m. C Moving positive q +2 n. C The field E does positive work, the P. E. decreases. If +2 n. C moves from A to C (closer to +Q), the field E does negative work and P. E. increases.

Example 2. What is the change in potential energy if a +2 n. C charge moves from A to B? Potential Energy: A · ·B 12 cm 8 cm From Ex-1: UA = + 1. 35 m. J DU = UB – UA = 0. 9 m. J – 1. 35 m. J +Q +6 m. C DU = -0. 450 m. J Note that P. E. has decreased as work is done by E.

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

Example 3. What is the change in potential energy if a -2 n. C charge moves from A to B? Potential Energy: From Ex-1: UA = -1. 35 m. J (Negative due to – charge) UB – UA = -0. 9 m. J – (-1. 35 m. J) A · 8 cm ·B 12 cm +Q +6 m. C DU = +0. 450 m. J A – charge moved away from a + charge gains P. E.

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

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

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

Example 4: Find the potential at a distance of 6 cm from a – 5 n. C charge. P. q = – 4 m. C r 6 cm -- Q = -5 n. C Negative V at Point P : VP = -750 V What would be the P. E. of a – 4 m. C charge placed at this point P? U = q. V = (-4 x 10 -6 m. C)(-750 V); U = 3. 00 m. J 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. Q 1 - r 1 r 3 Q 3 - ·A r 2 + Q 2 Potential is + or – based on sign of the charges Q.

Example 5: Two charges Q 1= +3 n. C and Q 2 = -5 n. C are separated by 8 cm. Calculate the electric potential at point A. B· 2 cm Q 1 + +3 n. C 6 cm A · 2 cm VA = 450 V – 2250 V; VA = -1800 V Q 2 = -5 n. C

Example 5 (Cont. ): Calculate the electric potential at point B for same charges. B· 2 cm Q 1 + +3 n. C 6 cm A · 2 cm VB = 1350 V – 450 V; VB = +900 V Q 2 = -5 n. C

Example 5 (Cont. ): Discuss meaning of the potentials just found for points A and B. Consider Point A: VA = -1800 V 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. B· 2 cm Q 1 + +3 n. C 6 cm A · 2 cm Q 2 = -5 n. C

Example 5 (Cont. ): Discuss meaning of the potentials just found for points A and B. Consider Point B: VB = +900 V B · 2 cm 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. Q 1 + +3 n. C 6 cm A · - 2 cm Q 2 = -5 n. C

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 Work. AB = 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. What work is done by the E-field if a +2 m. C charge is moved from A to B? B · 2 cm VA = -1800 V VB = +900 V Q 1 + +3 n. C VAB= VA – VB = -1800 V – 900 V VAB = -2700 V Note point B is at higher potential. 6 cm A · Q 2 - 2 cm -5 n. C Work. AB = q(VA – VB) = (2 x 10 -6 C )(-2700 V) Work = -5. 40 m. J E-field does negative work. Thus, an external force was required to move the charge.

Example 6 (Cont. ): Now suppose the +2 m. C charge is moved from back from B to A? B · 2 cm VA = -1800 V VB = +900 V Q 1 + +3 n. C VBA= VB – VA = 900 V – (-1800 V) VBA = +2700 V 6 cm A · 2 cm This path is from high to low potential. Q 2 - -5 n. C Work. BA = q(VB – VA) = (2 x 10 -6 C )(+2700 V) Work = +5. 40 m. J E-field does positive work. The work is done BY the E-field this time !

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

Example 7: The potential difference between two parallel plates is 800 V. If their separation is 3 mm, what is the field E? VA + + +q E F = q. E VB - - - 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: Work. AB = q(VA – VB) Oppositely Charged Parallel Plates: Work BY E-field

CONCLUSION: Chapter 25 Electric Potential