Electric Potential and Potential Energy Electric Potential Work-energy theorem: Change in potential energy = work done m Higher PE Gravitational Potential Energy It requires a certain amount of work to raise an object • Chapter 17 (Giancoli) of mass m from the ground to some distance above the • All sections except 17.6 (electric m Lower PE ground. dipoles) i.e. We have increased the potential energy of the object. PE=W=Force x Displacement Electric Potential Energy ⎛ kqQ ⎞ Find the work done in bringing a charge q from infinitely ∆ W = − F∆ r = −⎜ 2 ⎟⋅∆ r far away to a distance R from charge Q. ⎝ r ⎠ Q R R ∆r kqQ PE = ∆W = - ∆r F ∫ ∫ r 2 + +q ∞ ∞ R R ⎡1⎤ Charge is moved towards R = kqQ⎢ ⎥ by increments of ∆r ⎣ r ⎦∞ kqQ For a small displacement ∆r, the work done is: PE = *Note: PE = 0 when R ∞ R ∆W = force x displacement = -F • ∆r (we have a negative sign as the direction of the force is This is the PE of a charge q when it is a distance R from Q. opposite to the direction of the displacement) • PE is a scalar quantity If the PE is negative (when the charges have opposite signs), then the work is done by the charge, decreasing its PE. • The sign of the charges must be kept in all calculations Displacement + F q • Depending on the signs of Q and q, the PE could be positive or negative If the PE is positive (when the charges are both positive or both negative), then work must be done on the charge q to bring it closer to Q, increasing its PE. Displacement F q 1 Potential Energy of a System of Charges : Electric Potential The potential energy (PE) of a charge or a collection of Definition: Define the electric potential, V, at a point by the charges is the amount of work needed in assembling the PE of a charge at that point divided by the value of the charge. charges. Example: What is the PE of a charge Q located at point P? PE V = Units for V: J/C Q Q q P ∞ The potential, similar to PE, is a scalar quantity. i.e. We must keep the sign of the charges in all calculations PE is the work done in bringing the charge from ∞ to point P. Work done = force × displacement = 0 , as there is no force acting on the charge. PE = 0 For a point charge, Q´ The potential V at a point can be due to the presence of a ⎛ kqQ′ ⎞ 1 single charge or due to many charges. V = ⎜ ⎟ ⎝ r ⎠ q For many charges, Q1, Q2, Q3, ... , the potential at a point P is kQ′ V = given by the superposition principle: r P Vp = V1 + V2 + V3 +… r1 Q1 r3 kQ kQ kQ r2 = 1 + 2 + 3 +… This is the potential a distance r from the charge Q´. r1 r2 r3 Q Q * This property of the charge Q´ is more fundamental than PE. 2 3 * Remember to include the sign of the charges Q1, Q2, Q3, ... , If we know the potential at a point in space, when we place a when calculating Vp. charge Q at that point, its PE can be directly give by: * If a charge Q is placed at point P its potential energy will be: PE = QV PE = Q Vp Example : a) Calculate the potential at point P in the figure; b) What is the potential energy of a charge Q3=2 µC when placed at P? Q1 = 3 µC Q2 = –6 µC Q1 = 3 µCQ3= 2 µC Q2 = –6 µC P • r =6 cm r =8 cm • + 1 2 + r1=6 cm + r2=8 cm P PE = Q VP V = V + V P 1 2 = (2 × 10−6C)(−2.25 × 105J / C) kQ kQ = 1 + 2 = −0.45 J r1 r2 (9×109 N⋅m2 /C2)(3×10−6 C) (9×109 N⋅m2 /C2)(−6×10−6 C) Potential is not affected by the presence of a charge at that = + point. (Charges cannot contribute to their own potential) (0.06m) (0.08m) = −2.25×105 J /C If free to move, an object will always try to lower its PE. 2 Example: Along a line joining the two charges, Q1 = 40 µC Vp = V1 + V2 = 0 and Q2 = -60 µC, separated by 4 m, where does V vanish? kQ kQ 1 + 2 = 0 r1 r2 −6 2 −6 2 Q = 40 µC Q = –60 µC 40×10 N ⋅m /C 60×10 N ⋅m / C 1 2 ⇒ − = 0 P − • (x)m (4 x)m + r1= x r2= 4-x ⇒ 40(4 − x) = 60x Q2 has more charge than Q1,therefore P must be closer to Q1 (V ∼ Q/r) Solving for x, we get x = 1.6 m V = V + V = 0 Therefore, Q1 = 40 µC Q2 = –60 µC p 1 2 P • kQ1 kQ2 + r1= 1.6 m r2= 2.4 m + = 0 r1 r2 −6 2 −6 2 Are there other solutions? 40×10 N ⋅m /C 60×10 N ⋅m /C ⇒ − = 0 (x)m (x + 4)m To the left of Q1 ⇒ 40(x + 4) = 60x Q1 = 40 µC Q2 = –60 µC P + Solving for x, we get x = 8 m r 1= x r2= x + 4 Sometimes it is not necessary to know the absolute value of V or PE. Q = 40 µC Q = –60 µC 1 2 Example For convenience, we P + m can take the PE of the object of mass m r1= 8 m h r =12 m to be zero when it is 2 on the ground. ground Also, V= 0 at +ve infinity, and -ve infinity Therefore, the potential energy of a charge Q at P is equal to Then, at a height h above the ground, its PE, relative to the ground is mgh. PE (∞) = 0. 3 Example Similarly for the electric potential and potential energy, m sometimes we are interested in the changes in V and PE. h ∆ PE ∆ V = table q Where ∆V is the potential difference and ∆PE is the change in the potential energy, measured from a convenient reference ground point. In this example, we may choose to take the zero of PE at the top of the table. Then the PE of the object a distance h above the table is mgh. * The zero of PE is chosen for convenience. Experimentally, we always perform measurements in the Example: What is the PE of the two charges ? laboratory (and not at infinity!), so the symbol ‘∆’ is not really A B used, i.e. V ≡∆V. Q1 Q2 R Historically, the difference in potential was measured or calculated with the unit of volts [V] while the absolute Bring one charge at a time from ∞, calculating the work done in each potential was measured or calculated with the unit of joules case. The PE of the system of the two charges is the total work done. per coulomb [J/C]. The two units are equivalent and are now For the first charge, Q , the work done is zero, as V is zero, W = 0. used interchangeably. 1 A 1 For the second charge, Q2, the potential at point B is not zero. kQ 1V = 1 J/C 1 ⎛ kQ1 ⎞ kQ1Q2 VB = W2 = VBQ2 = ⎜ ⎟Q2 = R ⎝ R ⎠ R kQ Q PE = W + W = 1 2 1 2 R Note: This is also the PE of Q1 or Q2. Example : Consider three charges, Q1 = 2 µC, Q2 = – 1 µC 1. Start with Q1. The work done, W1, in bringing the charge and , Q3 = 3 µC located at points A, B and C as shown in the diagram. Calculate the PE of the three charges. from ∞ to point A is given by: W1 = Q1 VA = 0 {as VA = 0 (no charges are present )} AB C 2. Next bring Q2. The work done is W2 = Q2 VB. VB is not 2 m 3 m zero as Q1 is present at point A. Q1 Q2 Q3 AB 2 m Solution : Assume that the three charges are infinitely far Q1 Q2 away. Bring the charges, one at a time, each time calculating the work done on the charge. The total work done is the PE of W = Q V = Q (kQ /R) = – 0.009 J the system of the three charges (work-energy theorem). 2 2 B 2 1 4 3. For the last charge, Q3, the work done is W3 = Q3 VC, Example : Two charges, Q1=40 µC and Q2=-60 µC are 4 m where both Q1 and Q2 contribute to the potential at point C. apart. How much work is done in moving a charge Q=50 µC from point A to point B? Q = 40 µC Q = -60 µC AB C 1 2 2 mA 2 m 2 m 3 m + • Q1 Q2 Q3 2 m W3 = Q3 { (kQ1/R1) + (kQ2/R2) } = 0.0018 J B Total work done = W1 + W2 + W3 = – 0.0072 J Work done = change in potential energy ∴ PE = – 0.0072 J = QVB –QVA Work done = Q ( VB -VA) Q1 = 40µC Q2 = -60µC 2mA 2m + • = 50 x10-6 C(-6.39 x 104 J/C - (-9 x 104 J/C)) 2m r =2 2m = 1.3 J r1 =2 2m 2 B kQ1 kQ2 VB = + r1 r2 (9×109 N ⋅m2 /C2 )(40×10−6 C) (9×109 N ⋅m2 /C)(60×10−6 C) = − (2 2) m (2 2) m = −6.39×104 J /C Conservation of Energy Example : Two fixed charges, Q 1= 6 µC and Q2 = 4 µC are 4 m apart.