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Lecture 2: May 20th 2009 Physics for Scientists and Engineers II Physics for Scientists and Engineers II , Summer Semester 2009 Electric Field due to a Continuous Charge Distribution • • We can model a system of charges as being continuous (instead of discrete) if the distance between the charges is much smaller than the distance to the point where the electric field is calculated. Procedure: - Divide charge distribution into small charge elements Dq. - Add contributions to E from all charge elements. Dq D E ke 2 rˆ r Dq E ke 2 rˆi i ri r̂ DE r P Dqi dq ˆ E ke lim 2 ri ke 2 rˆ r ri D qi 0 i Physics for Scientists and Engineers II , Summer Semester 2009 Dq Charge Density (a useful concept when calculating E from charge distribution) Volume charge density (if Q is uniformly distribute d throughou t a volume V) : Surface charge density (if Q is uniformly distribute d on a surface of area A) : Linear charge density (if Q is uniformly distribute d on a line of length l) : Physics for Scientists and Engineers II , Summer Semester 2009 Q V Q A Q l dq dV dq dA dq dl Example: Electric Field due to a Uniformly Charged Rod dq = dx y dx x E x P a Contributi on to E from dq : l dE ke l a E from all dq (entire rod) : E a ke Notice : For l 0 E dq dx k e x2 x2 dx ke 2 ke x l a a keQ Q1 1 l a l a al a) keQ , the field of a point charge. 2 a Physics for Scientists and Engineers II , Summer Semester 2009 l a dx 1 k e 2 x x a Example: Electric Field due to a Uniformly Charged Rod…..this is harder…. y dE a dq = dx P r x x l Contributi on to E from dq : dx sin iˆ 2 r dx x ke 2 iˆ r r x dx ke 3 iˆ r x dx ˆ ke i 3 2 2 2 a x d E ke Physics for Scientists and Engineers II , Summer Semester 2009 dx cos ˆj 2 r dx a ˆ ke 2 j r r a dx ke 3 ˆj r a dx ˆ ke j 3 2 2 2 a x ke Example: Electric Field due to a Uniformly Charged Rod…..this is harder…. y dq = dx P dE a r x x l d E ke Contributi on to E from dq : a x dx 2 x2 3 2 iˆ ke a a dx 2 x2 3 ˆj 2 x dx ˆ a dx E ke i k e 3 3 2 2 2 2 2 2 0 a x a x l Total electric field at point P : l E x ke 0 a x l x 2 2 3 dx 2 Physics for Scientists and Engineers II , Summer Semester 2009 E y ke a 0 a 1 2 x 2 3 dx 2 ˆj ….solving the integral for Ex l E x ke a 0 x 2 x 2 3 dx 2 Substituti on : u a x l E x ke 0 2 2 1 1 x a2 x2 a2 x2 1 1 ke a2 l 2 a du 2 1 2 a x dx 2 dx ke x2 a 2 l 2 a 1 1 1 du k e u2 u a Q a 2 l 2 a ke l a a 2 l 2 Physics for Scientists and Engineers II , Summer Semester 2009 2 a 2 l 2 ….solving the integral for Ey l E y ke 0 a a 2 x 2 3 dx 2 Substituti on : x a tan E y ke max 0 k e a max 0 a dx a 2 a tan 2 2 1 1 2 cos 3 2 3 2 a d cos 2 a d cos 2 1 d cos 2 k e a ke sin 0 max keQ sin max a al kQ keQ l e a l l 2 a2 a l 2 a2 (I wouldn' t expect you to know that) ke a max 0 1 1 tan 2 3 2 1 d cos 2 max cos d 0 a max l 2 a2 l sin max Physics for Scientists and Engineers II , Summer Semester 2009 l l 2 a2 ….and the final result Q a 2 l 2 a E x ke l a a 2 l 2 In the limit lim l 0 lim l keQ a l 2 a2 of a very short rod : : Ex 0 In the limit Ey and E y keQ a2 (field of a point charge again) of a very long rod : : E x ke a and E y ke Physics for Scientists and Engineers II , Summer Semester 2009 a Visualizing Electric Fields with Electric Field Lines • • • The electric field vector is always tangent to the electric field line. The electric field line has a direction (indicated by an arrow). The direction is the same as that of the electric field (same direction as force on a positive test charge). The number of lines per unit area through a normal plane (perpendicular to field lines) is proportional to the magnitude of the electric field in that region. Example: Electric field lines of a point charge N field lines + Surface density of field lines at an N imagined sphere of radius r is 4 r 2 Electric field strength is proportional to Physics for Scientists and Engineers II , Summer Semester 2009 1 r2 Visualizing Electric Fields with Electric Field Lines • • • • • For a single positive point charge: Electric field lines go from the positive charge to infinity. For a single negative point charge: Electric field lines go come from infinity and end at the negative point charge. For multiple point charges: Lines can start at the positive charges and end at the negative charges. Electric field lines can never cross (think about why that is so). For two unequal point charges of opposite sign with charges Q1 and Q2 , the number N1 of field lines terminating at Q1 and the number N2 of field lines terminating at Q2 are related by the equation N 2 Q2 N1 Q1 Physics for Scientists and Engineers II , Summer Semester 2009 Motion of a Charged Particle in a Uniform Electric Field • • Assume particle has charge q, mass m. Particle experiences a force • The force results in an acceleration (according to Newton’s second law): Fe q E a Fe qE m m • • • For positive charges: Acceleration is in the same direction as electric field. For negative charges: Acceleration is in a direction opposite to the electric field. A uniform electric field will cause a constant acceleration of the particle. You can use equations of motion for constant acceleration. • Work is done on the particle by the electric force as the particle moves. W Fe D x Physics for Scientists and Engineers II , Summer Semester 2009 Example (similar to Ex. 23.10 in book) + + + - vi ? + + + + + + + E - - - - - - - - - L = 0.100 m Electron: m = 9.11x10-31 kg ; q = 1.60x10-19 C Electric Field: E = 800 N/C The electron leaves the electric field at an angle of = 65 degrees. Q1: What was the initial velocity of the electron? Q2: What is the final velocity of the electron (magnitude)? Q3: How low would the electric field have to be so that the net force on the electron is zero? Q4: Were we justified in neglecting the gravitational force in Q1 and Q2? Physics for Scientists and Engineers II , Summer Semester 2009 Question 1 : vy f F qE ay t e t t m m vy f qE L m vi L t vi vi tan ; vy f vi tan qE L m vi N 1.60 10 19 C 800 0.100m qE L C 6 m vi 2 . 5 10 m tan 9.10 10 31 kg tan 65 deg s Question 2 : 2.5 106 m vi s 6.1106 m vf cos cos65 s Physics for Scientists and Engineers II , Summer Semester 2009 Question 3 : qE m g E gm q 9.8 m 2 9.10 10 31 kg s 1.6 10 19 C 5.6 10 11 N C Question 4 : Yes, except for extremely small electric fields, the electric force on an electron is much larger tha n the gravitatio nal force. Physics for Scientists and Engineers II , Summer Semester 2009 Gauss’s Law – An alternative procedure to calculate electric fields of highly symmetric charge distributions The concept of “Electric Flux”: Area = A E Electric flux : E A E for constant E and E being perpendicu lar to area. Physics for Scientists and Engineers II , Summer Semester 2009 Area perpendicu lar to E is A A A cos E Area not perpendicu lar to E is A The electric flux through the two surfaces is the same E E A E A cos Physics for Scientists and Engineers II , Summer Semester 2009 Normal to green surface The electric flux through the two surfaces is the same Normal to green surface E E A E A cos To calculate the flux through a randomly oriented area you need to know the angle between the electric field and the normal to the area. Physics for Scientists and Engineers II , Summer Semester 2009 How to treat situations where the electric field is not constant over the area? • Divide area into small areas over which E is constant. • Calculate flux for each small area. • Add fluxes up. Ei Ai i Area vector: magnitude = area direction = perpendicular to area Electric flux throu gh surface element : D E Ei DAi cos i E i D Ai Electric flux throu gh entire surface : E E i D Ai i ....and in the limit of infinitesi mally small surface segments : E Ed A surface Physics for Scientists and Engineers II , Summer Semester 2009 “surface integral” Flux through a closed surface: •Convention: Area vectors always point outwards. Field lines that cross from the inside to the outside of the surface : 90 (positive flux because cos is positive) Field lines that cross from the outside to the inside of the surface: 90 180 (negative flux because cos is negative) Flux throu gh closed surface : E E d A En dA Physics for Scientists and Engineers II , Summer Semester 2009 Example: Cube in a uniform field dA3 E dA1 dA6 dA2 dA5 dA4 E E d A E d A E d A E d A E d A E d A E d A 1 2 3 E d A E d A 1 4 0 5 0 6 0 0 2 E cos180dA E cos 0dA E dA E dA EL2 EL2 0 1 2 Physics for Scientists and Engineers II , Summer Semester 2009 1 2