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# Electrostatics & Electric Fields

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A set of slides created to teach G11 Electrostatics & Electric Fields P to learners at Bishops Diocesan College in Cape Town.

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### Electrostatics & Electric Fields

1. 1. Electrostatics & Electric Fields K Warne SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net
2. 2. Coulomb’s Law: F = kQ1Q2 r2 F Q2 The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Q1 k = coulomb’s constant = 9 x 109 N. m2.C-2 Calculate the force between an electron and a proton if the distance between them is 1nm. (e- = -1.6x10-19 ) F = kQ1Q2/r2 = (9x109)(-1.6x10-19 )(1.6x10-19)/(1x10-9)2 = -2.304x10-10N
3. 3. Increasing the charge on any one of the spheres will increase the force by a proportion al amount. F = kQ1Q2 r2 F Q2Q1 1. F2= k2Q1Q2 r2 2F Q2 2Q12. F2=2F SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
4. 4. 7.1 ELECTRIC FIELD A region ____________ in which a charge will experience a “_________” or electrostatic _______________________. + -+ + - ELECTRIC FIELD LINE: A line drawn in such a way that at at any point on the line, a small ___________ charge will experience a ___________ in the direction of the ______________ to that line. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
5. 5. 7.2 ELECTRIC FIELD PATTERNS: - + ++ VERY SMALL POINT CHARGES NEAR ONE ANOTHER - + SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
6. 6. + + + + + + + + + + + - - - - - - - - - - - Between oppositely charged plates Field is _______________ the oppositely charged plates. (Force experienced by a charge placed anywhere between the plates is ___________________) SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
7. 7. ELECTRIC FIELD STRENGTH (E) For an electric field E = / where E = ______________ strength in ______ F = _______ in___ Q = _________ in ___ NB. Electric field strength is a ___________quantity (direction: _____________ to ____________) + F E Eg: What force would be experienced by an electron in an electric field of 1 x 10-6 NC-1? SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
8. 8. qF E Q E = F q but F = kQ1Q2 r2 so E = q kQq r2 so E = kQ r2 r Qq SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
9. 9. Electric field due to multiple charges  Calculate the electric field strength at X. X B -2C A +2C 1cm2cm Field due to A: E = kQ1 /r2 =(9x109)(2x10-6)/(2x10-2)2 = 4.5 x 107 N.C-1 away from A Field due to B: E = kQ1 /r2 =(9x109)(-2x10-6)/(1x10-2)2 = -1.8x108 N.C-1 = 1.8x108 N.C-1 towards B E = E1 + E 2 = 4.5 x 107 + ( 1.8x108 ) = 2.25x108 N.C-1 E = 2.25x108 N.C-1 TOWARDS B (AWAY FROM A)SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
10. 10. Work done W = F x d but F = QE (Def of E)  W = QEd If the charge is pushed to the left the work done on the charge is: W = QEd If the charge is now released, it moves spontaneously to the right,because the field does work on the charge: W = QEd Kinetic energy gained = work done 1/2mv2 = QEd => v = √2QEd/m Consider applying a force F needed to move a charge from A to B. The charge moves a distance d. The size of the charge is Q. 7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD + d - + - + - + B A - + - + - + - + - + - + - + - SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
11. 11. Question  A proton is accelerated by an electric field of 1.5x106 N.C-1 over a distance of 2 nm. The mass of a proton is 1.7 x 10-27 kg  Calculate the final velocity attained by the proton if it started from rest. E = 1.5x106 N.C-1 d = 2 nm = 2 x 10-9 m m = 1.7 x 10-27 kg Q = 1.6 x 10-19 C v = ? Formula? v = √2QEd/m SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
12. 12. • A positive charge moves spontaneously … ............................................... of the field. • A …………………………….. moves spontaneously in a direction …………… to that of the electric field. • Thus, at any point in an electric field an electric charge possesses …………………… (………..) • Where free to move, it will ……………….. • It will therefore gain ………. as it loses ……. ie. …………… = ………………. (ignoring air friction). ……………………… = …………………… 7.5 WORK, Ep AND Ek IN ELECTRIC FIELDS + - + - SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
13. 13. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnesscience.net Energy in a radial field  Work must be done on Q2 to move it across a distance r from Q1.  If released Q2 must therefore have the potential to move away again with that same energy.  Q2 therefore has Potential energy U at a distance r from Q1. + Q2 Q1 •r Energy = Work done = F x d = kQ1Q2 x r r2 U = kQ1Q2 r
14. 14. A positive test charge  is at a ……………….. …………………. Ep at B and at a ………… at A. There is thus a potential difference (V) between B and A. POTENTIAL DIFFERENCE (V) +   B A Potential difference = Unit: …………………………………. The potential difference between two points in an electric field is the ………………………………………………………. in moving the charge from the one point to the other. Define the electric potential at a point as the …………………………… per ………………, i.e. the potential energy …………………………. would have if it were placed …………………... SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
15. 15. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnesscience.net Electric Potential + Q2 Q1 •r = kQ1Q2 r Q2 = kQ1 r Charge at that point
16. 16. + - + - + - + - + - + - + - + - ELECTRIC FIELD BETWEEN 2 PARALLEL PLATES If plates are s apart and p.d across plates is V, then V = W = QEd = Ed Q Q E = V d where E = electric field strength in Vm-1 V = potential difference in V and d = distance apart in m NOTE: V.m-1 = N.C-1 [V.m-1 = J.C-1.m-1 = N.m.C-1m-1 = N.C-1] V d SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
17. 17. Milikan’s Experiment • These droplets are charged as they are forced out of the nozzle. • Milikan used a microscope to observe the oil droplets between the plates. • As the plate voltage increases some of the drops fall more and more slowly until the drops stop moving. • At this point the electric force is equal to the weight of the oil droplet. V = Ed = (F/q).d .: F = q.V/d Felec = Fg qV = mg d q=Droplet Charge V= Holding Voltage d= Distance between plates m=droplet mass g= Acceleration due to gravity. • From his experiments Milikan determined that the charge on an electron was 1.6×10-19 C. • By timing how long it takes for a droplet to fall with the plates switched off he could calculate the mass of the droplet. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> ScienceCafe
18. 18. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnesscience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience
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