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Example Problem A particle with charge +2 nC (1 nanoCoulomb=10-9 C) is located at the origin. What is the electric field due to this particle at a location <-0.2,-0.2,-0.2> m? Solution: 1. Distance and direction: r 1 𝑞1 𝐸1 = 𝑟 2 4𝜋𝜀0 𝑟 𝑟 = 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑_𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 − 𝑠𝑜𝑢𝑟𝑐𝑒_𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑟 = −0.2, −0.2, −0.2 − 0, 0, 0 𝑟 = (−0.2)2 + (−0.2)2 + (−0.2)2 = 0.35m 𝑟 −0.2, −0.2, −0.2 𝑟= = = −0.57, −0.57, −0.57 𝑟 0.35 Example Problem 2. The magnitude of the electric field: 1 𝑞1 𝐸1 = 𝑟 2 4𝜋𝜀0 𝑟 2 2 × 10−9 C Nm 1 𝑞 N 9 = 9 × 10 𝐸 = = 147 2 2 2 C 0.35 m 4𝜋𝜀0 𝑟 C 𝑦 3. The electric field in vector form: N −0.57, −0.57, −0.57 𝐸 = 𝐸 𝑟 = 147 C 𝑥 N 𝐸 = −84, −84, −84 C 𝑧 𝐸 Clicker Question 1 A penny carrying a small amount of positive charge Qp exerts an electric force F on a nickel carrying a large amount of positive charge Qn that is a distance d away (Qn > Qp ). Which one of the following is not true? A. The electric force exerted on the penny by the nickel is also equal to F. B. The number of electrons in the penny is less than the number of protons in the penny. C. 𝑭 ≈ 𝟏 𝑸𝒑 𝑸𝒏 , 𝟒𝝅𝜺𝟎 𝒅𝟐 if d is small compared to the size of the coins. D. 𝑭 ≈ 𝟏 𝑸𝒑 𝑸𝒏 , 𝟒𝝅𝜺𝟎 𝒅𝟐 coins. if d is large compared to the size of the Clicker Question 2 A positive and a negative charge are separated by a distance r, r +q1 What are the directions of the forces on the charges? Choice 𝑭 𝒐𝒏 𝒒𝟏 𝑭 𝒐𝒏 𝒒𝟐 A Left Left B Right Left C Left Right D Right Right -q2 What is the magnitude of the self-force? Choice A B C 𝑭 𝒐𝒏 𝒒𝟐 𝒅𝒖𝒆 𝒕𝒐 𝒒𝟐 infinite k𝒒𝟏 𝒒𝟐 /𝒓𝟐 0 The Coulomb Force 1 Q1Q2 F r̂ 2 4 0 r 1 Q1Q2 F F 2 40 r 0 = permittivity constant r + + 2 1 Force repulsive F21 Force on “2” by “1” + r - F21 2 1 Force attractive • The force exerted by one point charge on another acts along line joining the charges. • The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs. How Strong is the Coulomb Force Electric Field E F /q E E x, y , z , t Electric field has units of Newton per Coulomb: [N/C] No ‘self-force’! q E 2 40 r 1 r 0, E Point charge does not exert field on itself! The Superposition Principle +q2 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 3 -q1 +q3 𝐸1 𝐸 𝑛𝑒𝑡 𝐸2 The net electric field at a location in space is a vector sum of the individual electric fields contributed by all charged particles located elsewhere. The electric field contributed by a charged particle is unaffected by the presence of other charged particles. The Superposition Principle +q2 +q3 -q1 𝐸1 𝑃 𝐸2 𝐸 𝑛𝑒𝑡 𝐸 3 The E of a Uniformly Charged Sphere Can calculate using principle of superposition: 1 Q Esphere rˆ 2 40 r Esphere 0 Recall this every night before bed! for r>R (outside) for r<R (inside) The Superposition Principle The electric field of a dipole: Electric dipole: Two equally but oppositely charged point-like objects s -q +q Example of electric dipole: HCl molecule What is the E field far from the dipole (r>s)? Calculating Electric Field Choice of the origin y s -q +q x z Choice of origin: use symmetry s 1. E along the x-axis 𝐸−,𝑥 -q +q E1, x E , x E , x E1, x E1, x 𝐸1,𝑥 𝐸+,𝑥 r 1 q 4 0 r s 2 2 1 q 4 0 r s 2 2 1 qr 2 qrs qs 2 / 4 qr 2 qrs qs 2 / 4 4 0 1 r s 2 r s 2 2 2qrs 4 0 r s 2 2 r s 2 2 2 Approximation: Far from the Dipole E1, x 1 2 srq 2 2 40 s s r r 2 2 2 2 s s if r>>s, then r r r 2 2 2 E1, x 1 2 sq 40 r 3 E1 1 2sq ,0,0 3 40 r While the electric field of a point charge is proportional to 1/r2, the electric field created by several charges may have a different distance dependence. 2. E along the y-axis 𝐸+ E 𝐸2 1 q rˆ 2 40 r s s r 0, y ,0 ,0,0 , y ,0 2 2 s s r 0, y ,0 ,0,0 , y ,0 2 2 𝐸− y 𝑟+ s y2 2 -q s +q 2 E E 1 q 40 s y2 2 1 q 40 s y2 2 r r s , y ,0 2 2 s y2 2 2 s , y ,0 2 2 s y2 2 2 s y2 2 2 2. E along the y-axis 𝐸+ E 𝐸2 𝐸− y 𝑟+ 1 q 40 s y2 2 s , y ,0 2 2 s y2 2 2 1 E2 E E 40 E 1 q 40 s y2 2 qs q 33 2 2 2 s y 2 s +q 2 s y2 2 1s,0,0 1 qs ,0,0 if r>>s, then E2 3 4 0 r -q s , y ,0 2 at <0,r,0> 2 3. E along the z-axis y E1 E2 1 qs ,0,0 3 40 r at <0, r, 0> or <0, 0, r> z 1 2sq ,0,0 3 40 r at <r, 0, 0> x Due to the symmetry E along the z-axis must be the same as E along the y-axis! Other Locations The Electric Field Point Charge: E1 1 q1 rˆ 2 40 r Dipole: for r>>s : y +q z 1 2qs E ,0,0 3 4 0 r x - at <r,0,0> 1 qs ,0,0 3 4 0 r at <0,r,0> 1 qs E ,0,0 3 4 0 r at <0,0,r> E s -q + Example Problem y A dipole is located at the origin, and is composed of E=? particles with charges e and –e, separated by a distance 210-10 m along the x-axis. Calculate the magnitude of the E field at <0, 210-8, 0> m. 1 sq Since r>>s: E1, x 200Å 40 r 3 2 10 19 Nm 2 10 m 1.6 10 C 9 E1, x 9 10 3 2 C 2 10 8 m 4 N x E1, x 7.2 10 C 2Å Using exact solution: 4 N E1, x 7.1999973 10 C Interaction of a Point Charge and a Dipole +q Edipole +Q -q s F 𝑑≫𝑠 F QEdipole Q F 1 2qs ,0,0 3 40 d • Direction makes sense? - negative end of dipole is closer, so its net contribution is larger • What is the force exerted on the dipole by the point charge? - Newton’s third law: equal but opposite sign Dipole Moment x: y, z: E1 E2 1 2qs p ,0 ,0,0 ,0 33 40 rr r>>s 1 qs p ,,00,,00 33 40 r The electric field of a dipole is proportional to the Dipole moment: p = qs p qs, direction from –q to +q Dipole moment is a vector pointing from negative to positive charge Dipole in a Uniform Field F qE Forces on +q and –q have the same magnitude but opposite direction Fnet qE qE 0 It would experience a torque about its center of mass. What is the equilibrium position? Electric dipole can be used to measure the direction of electric field. Choice of System Multiparticle systems: Split into objects to include into system and objects to be considered as external. To use field concept instead of Coulomb’s law we split the Universe into two parts: • the charges that are the sources of the field • the charge that is affected by that field A Fundamental Rationale • Convenience: know E at some r location – know the electric force on any charge: F qE • Can describe the electric properties of matter in terms of electric field – independent of how this field was produced. Example: if E>3106 N/C air becomes conductor • Retardation Nothing can move faster than light c c = 300,000 km/s = 30 cm/ns Coulomb’s law is not completely correct – it does not contain time t nor speed of light c. F 1 q1q2 rˆ 2 40 r E 1 q rˆ 2 40 r v<<c !!!