2.
Coulomb’s Law: F Q 2 The electrostatic force between two point charges is ...........................to the product of the charges and ................ proportional to the ........................of the distance between them. Q 1 k = .................. constant = 9 x 10 9 .... . .... 2 . .... -2 Calculate the force between an electron and a proton if the distance between them is 1nm. F = k … 1 ….. 2 r …
3.
Coulomb’s Law: F Q 2 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. Q 1 k = coulomb’s constant = 9 x 10 9 N.m 2 .c -2 Calculate the force between an electron and a proton if the distance between them is 1nm. F = k Q 1 Q 2 r 2
4.
Increasing the charge on any one of the spheres will ………………… the force by a ………………… amount. Force and charge F Q 2 Q 1 1. F 2 = … . F Q 2 … Q 1 2. F 2 =…… F = k Q 1 Q 2 r 2
5.
Increasing the charge on any one of the spheres will increase the force by a proportional amount. Force and charge F Q 2 Q 1 1. F 2 = k 2 Q 1 Q 2 r 2 2 F Q 2 2 Q 1 2. F 2 = 2 F F = k Q 1 Q 2 r 2
6.
Increasing the ……………… between the spheres will ………………… the force by the …………………… of the proportional amount . Force and charge F 2 = 1 / 4 F Q 2 Q 1 2. F = k Q 1 Q 2 r 2 F Q 2 Q 1 1. … .. F 2 =………
7.
Increasing the distance between the spheres will decrease the force by the SQUARE of the proportional amount . Force and charge 1 / 4 F Q 2 Q 1 2. F = k Q 1 Q 2 r 2 F Q 2 Q 1 1. 2r F 2 = 1 / 4 F F 2 = k Q 1 Q 2 (2 r) 2
8.
7.1 ELECTRIC FIELD <ul><li>A region ____________ in which a charge will experience a “_________” or electrostatic </li></ul><ul><li>_______________________. </li></ul>+ - + + - 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.
9.
7.1 ELECTRIC FIELD <ul><li>A region in space in which a charge will experience a “coulomb” or electrostatic force of attraction or repulsion. </li></ul>This would occur at any point between the plates Repelled by positive plate - attracted by negative plate. + - + + - ELECTRIC FIELD LINE: A line drawn in such a way that at any point on the line, a small positive charge will experience a force in the direction of the tangent to that line. and so we can represent the field by a series of lines.
10.
PROPERTIES OF ELECTRIC FIELD LINES <ul><li>Stronger the field - _____________ field lines </li></ul><ul><li>Equally spaced parallel lines - ________ field. </li></ul><ul><li>(i.e. same force experienced at all points) </li></ul><ul><li>Field lines meet an object at _______________ </li></ul><ul><li>The field is continuous and _____________ </li></ul><ul><li>Field lines never ______ or ___________ because they represent _____________ forces. </li></ul>+ -
11.
PROPERTIES OF ELECTRIC FIELD LINES <ul><li>Stronger the field - closer field lines </li></ul><ul><li>Equally spaced parallel lines - uniform field </li></ul><ul><li>(i.e. same force experienced at all points) </li></ul><ul><li>Field lines meet an object at 90 </li></ul><ul><li>The field is continuous and is 3 dimensional. </li></ul><ul><li>Field lines never cross or intersect because they represent resultant forces. </li></ul>+ -
12.
7.2 ELECTRIC FIELD PATTERNS: - + - + + + VERY SMALL POINT CHARGES NEAR ONE ANOTHER
13.
7.2 ELECTRIC FIELD PATTERNS: VERY SMALL POINT CHARGES NEAR ONE ANOTHER - + - + + +
14.
Between oppositely charged plates + + + + + + + + + + + - - - - - - - - - - - Field is _______________ the oppositely charged plates. ( Force experienced by a charge placed anywhere between the plates is ___________________ )
15.
Between oppositely charged plates + + + + + + + + + + + - - - - - - - - - - - Field is uniform between the oppositely charged plates. ( Force experienced by a charge placed anywhere between the plates is identical )
16.
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 ?
17.
ELECTRIC FIELD STRENGTH (E) For an electric field E = F / Q where E = electric field strength in NC -1 F = force in N Q = charge in C NB. Electric field strength is a vector quantity (direction positive to negative) + F E Eg: The force experienced by an electron in an electric field of 1 x 10 -6 NC -1 ? F = EQ = ( 1 x 10 -6 )( 1.6 x 10 -19 ) =
18.
FIELD AROUND A SMALL CHARGED OBJECT Strength of field at A is: E = where: E = ………………………………… k = …………... constant = ………………… Q = ……. charge [C] r = …………………….. Example: Find the electric field strength 2 cm from a proton. + A …………………………………………………………………… …………………………………………………………………… Q • r
19.
FIELD AROUND A SMALL CHARGED OBJECT Strength of field at A is: E = k Q r 2 where: E = electric field strength at A k = coulomb’s constant = 9 x 10 9 N.m 2 .c -2 Q = + charge [C] r = distance (m) to A Example: Find the electric field strength 2 cm from a proton. E = kQ = 9x10 9 x 1,6 x 10 -19 = 3,6 x 10 -6 N.C -1 away r 2 (2 x 10 -2 ) 2 + A Q • r
20.
Electric field due to multiple charges <ul><li>Calculate the electric field strength at X. </li></ul>X B -2 C A +2 C 1cm 2cm Field due to A: E = Field due to B: E =
21.
Electric field due to multiple charges <ul><li>Calculate the electric field strength at X. </li></ul>X B -2 C A +2 C 1cm 2cm Field due to A: E = kQ/r 2 = Field due to B: E = kQ/r 2 = 2.25x10 8 N.C
22.
7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD Work done W = ……….. but F = ………. W = ………. If the charge is pushed to the left the …………………. on the charge is: W = ………… If the charge is now released, it moves spontaneously to the …………, because the …………… does …………………. on the charge : W = QEs Kinetic energy gained = ………………. …………………………………… . Consider applying a force F needed to …………………………………. The charge moves a ………………………… The size of the charge is …….. + s - + - + - + B A - + - + - + - + - + - + - + -
23.
7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD Work done W = F x s but F = QE W = QEs If the charge is pushed to the left the work done on the charge is: W = QEs If the charge is now released, it moves spontaneously to the right,because the field does work on the charge : W = QEs Kinetic energy gained = work done 1 / 2 mv 2 = QEs => v = √2QEs/m Consider applying a force F needed to move a charge from A to B . The charge moves a distance s . The size of the charge is Q . + s - + - + - + B A - + - + - + - + - + - + - + -
24.
MOVING at 90 0 to A UNIFORM ELECTRIC FIELD If a charge is moved at …….. to the field from A to B, + The force needed ……………………………… is ………. N . Therefore …………………….. ! ……………… is done on the charge! + A - + - + - + - + - + - + - + - + B -
25.
MOVING at 90 0 to A UNIFORM ELECTRIC FIELD If a charge is moved at 90 to the field from A to B, + + The force needed in the direction of motion is 0 N . Therefore no work done! no work is done on the charge! + A - + - + - + - + - + - + - + - + B -
26.
Question <ul><li>A proton is accelerated by an electric field of 1.5x10 6 N.C -1 over a distance of 2 nm. The mass of a proton is 1.7 x 10 -27 kg </li></ul><ul><li>Calculate the final velocity attained by the proton if it started from rest. </li></ul><ul><li>……………………………………………………………………………………………… . </li></ul><ul><li>……………………………………………………………………………………………… . </li></ul><ul><li>……………………………………………………………………………………………… . </li></ul><ul><li>……………………………………………………………………………………………… . </li></ul><ul><li>……………………………………………………………………………………………… . </li></ul>
27.
Question <ul><li>A proton is accelerated by an electric field of 1.5x10 6 N.C -1 over a distance of 2 nm. The mass of a proton is 1.7 x 10 -27 kg </li></ul><ul><li>Calculate the final velocity attained by the proton if it started from rest. </li></ul><ul><li>Kinetic energy gained = work done </li></ul><ul><li> 1/2mv 2 = QEs </li></ul><ul><li> v = √ 2QEs </li></ul><ul><li> m </li></ul><ul><li> = 2(1.6x10 -19 ) (1.5x10 6 )(2x10 -9 ) </li></ul><ul><li> 1.7x10 -27 </li></ul><ul><li> = 751.5 m.s -1 </li></ul>√
28.
7.5 WORK, E p AND E k IN ELECTRIC FIELDS <ul><li>A positive charge moves spontaneously … ............................................... of the field. </li></ul><ul><li>A …………………………….. moves spontaneously in a direction …………… to that of the electric field. </li></ul><ul><li>Thus, at any point in an electric field an electric charge possesses …………………… (………..) </li></ul><ul><li>Where free to move, it will ………………. . </li></ul><ul><li>It will therefore gain ………. as it loses ……. </li></ul><ul><li>ie. …………… = ………………. (ignoring air friction). </li></ul><ul><li>……………………… = …………………… </li></ul>+ - + -
29.
7.5 WORK, E p AND E k IN ELECTRIC FIELDS <ul><li>A positive charge moves spontaneously in the direction of the field. </li></ul><ul><li>A negative charge moves spontaneously in a direction opposite to that of the electric field. </li></ul><ul><li>Thus, at any point in an electric field an electric charge possesses potential energy (E p ) </li></ul><ul><li>This Electrical Potential energy is therefore due to the position of the charge relative to other charges. </li></ul><ul><li>Where free to move, it will accelerate . </li></ul><ul><li>It will therefore gain E k as it loses E p </li></ul><ul><li>ie. E k gained = E p lost (ignoring air friction). </li></ul>+ - + - Define the electrical potential energy of a charge as the energy it has because of its position relative to other charges that it interacts with
30.
Energy in a radial field <ul><li>Work must be done on Q 2 to move it to a distance r from Q 1 . </li></ul><ul><li>If released Q 2 must therefore have the potential to move away again with that same energy. </li></ul><ul><li>Q 2 therefore has Potential energy U at a distance r from Q 1 . </li></ul>+ Q 2 Q 1 • r Energy = Work done = F x s = kQ 1 Q 2 x r r 2 U = kQ 1 Q 2 r
31.
POTENTIAL DIFFERENCE (V) A positive test charge is at a ……………….. …………………. E p at B and at a ………… at A. There is thus a potential difference (V) between B and A. 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 …………………... +
32.
POTENTIAL DIFFERENCE (V) A positive test charge is at a higher potential energy E p at B and at a lower E p at A. There is thus a potential difference (V) between B and A. B A Potential difference = Potential lost/work done V = W charge moved Q Unit: volt I V =1 J.C -1 The potential difference between two points in an electric field is the work done per coulomb of charge in moving the charge from the one point to the other. Define the electric potential at a point as the electrical potential energy per unit charge , i.e. the potential energy a positive test charge would have if it were placed at that point. +
33.
THE VOLT <ul><li>NOTE: Zero E p exists at an infinite distance away from the charge which is producing the electric field. Absolute E p values are thus all negative. </li></ul>Definition: The volt: the potential difference between any two points in an electric field is one volt if one joule of work is done in moving one coulomb of charge from one point to the other.
34.
ELECTRIC FIELD BETWEEN 2 PARALLEL PLATES <ul><li>If plates are s apart and p.d across plates is V , then </li></ul><ul><li>V = W = QEs = Es </li></ul><ul><li>Q Q </li></ul><ul><li>E = V </li></ul><ul><li>s where </li></ul><ul><li>E = electric field strength in Vm -1 </li></ul><ul><li>V = potential difference in V and </li></ul><ul><li>s = distance apart in m </li></ul><ul><li>NOTE: V.m -1 = N.C -1 </li></ul><ul><li>[V.m -1 = J.C -1 .m -1 = N.m.C -1 m -1 = N.C -1 ] </li></ul>V s + - + - + - + - + - + - + - + -
35.
Milikan’s Experiment <ul><li>Milikan used a microscope to observe the oil droplets between the plates. </li></ul><ul><li>These droplets are charged as they are forced out of the nozzle. </li></ul><ul><li>As the plate voltage increases some of the drops fall more and more slowly until the drops stop moving. </li></ul><ul><li>At this point the electric fore is equal to the weight of the oil droplet. </li></ul><ul><li>The electric force on the object is equal to the weight. </li></ul><ul><li>qV = mg </li></ul><ul><li> d </li></ul><ul><li>q=Droplet Charge V= Holding Voltage d= Distance between plates m=droplet mass g= Acceleration due to gravity. </li></ul><ul><li>From his experiments Milikan determined that the charge on an electron was 1.6×10-19 C. </li></ul><ul><li>By timing how long it takes for a droplet to fall with the plates switched off he could calculate the mass of the droplet. </li></ul>
A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later.
Be the first to comment