Objects usually are charged up through the transfer of electrons from one object to the other.
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1 1 Pat 1: Electic Foce 1.1: Review of Vectos Review you vectos! You should know how to convet fom pola fom to component fom and vice vesa add and subtact vectos multiply vectos by scalas Find the esultant vecto R = whee 1 = 10 m, 45 and 2 = 20 m, 270. Ans m, Chage Symbol q, mks units [Coulomb = C] Fundamental chage e = C Chage of poton q p = +e Chage of electon q e = -e An object is neutal if it has the same numbe of electons and potons. An object has a net positive chage if it has moe potons than electons. An object has a net negative chage if it has moe electons than potons. Objects usually ae chaged up though the tansfe of electons fom one object to the othe. An object has a net chage of -8 nc. Does it have moe potons o electons? How many moe? Ans. The object has 50 billion moe electons than potons.
2 2 1.3 Electic Foce q 2 Coulomb s Law descibes the electic foce between chaged objects. Size: F = k q 1q 2 2 q 1 Coulomb s constant k = N m 2 /C 2 Opposite chages attact along line. Like chages epel along line. Conside the gound state of a H-1 hydogen atom that has one electon obiting a single poton in the lowest obit allowed fo the electon. Compae the size of the electic foce between the electon and poton to that of the gavitational foce. You will need the following infomation. mass of electon kg mass of poton kg distance between electon and poton in gound state 0.05 nm Ae we justified in saying that it is the electic foce that holds atoms togethe and that the gavitational foce is negligible in compaison? Ans. electic foce ~ N, gavitational foce ~ N. The electic foce is lage by 39 odes of magnitude. We can definitely ignoe gavity. The following thee chages ae held fied on an -y gid. q 1 = -2 C at (0,0) q 2 = +3 C at (2 m,0) q 3 = -4 C at (1 m, -2 m) Find the net foce eeted by chages 2 and 3 on chage 1. Ans N, 61.3
3 3 1.4 Electic Field Symbol mks units E [N/C] A chaged object with chage q o poduces an electic field vecto E at evey point in space ecept at its position. This object eets an electical foce on anothe chaged object with chage q. This foce is given by F = qe whee E is the electic field of q o at the position of q. If q is +, the foce points in same diection as the field vecto. If q is -, the foce points in the opposite diection as the field vecto. Point Chages A chaged object that can be appoimated as a point chage poduces an electic field at point P. Size of field: E = k q 2 Diection of field: If q is +, field vecto points away fom chage along line. +q If q is -, field vecto points towads chage along line. The following thee chages ae held fied on an -y gid. -q P P q 1 = -2 C at (0,0) q 2 = +3 C at (2 m,0) q 3 = -4 C at (1 m, -2 m) Note that this is the same chage distibution used in the pevious eample in the foce section. (a) Find the net electic field at the oigin due to these point chages. (b) Find foce eeted by chages 2 and 3 on chage 1 by using F = qe and show that this answe is identical to the one you found in the pevious eample. Ans. (a) 7344 N/C, (b) N, 61.3
4 4 Continuous Chage Distibutions If point P is nea a lage chaged object, then to find the electic field at P you chop up the object into many point chages. You find the electic field due to each point chage and sum up all these field vectos. Eample 1: Field on the cental ais of a unifomly chaged od of length L and chage E = k d(d + L) L If is +, field vecto points away fom od along ais. If is -, field vecto points towads od along ais. d P Eample 2: Field on the cental ais of a unifomly chaged loop of adius a and chage k E = ( 2 + a 2 ) 3/2 a P If is +, field vecto points away fom loop along ais. If is -, field vecto points towads loop along ais. Gauss s Law Gauss s Law is anothe way to find the electic field nea a lage chaged object. Gauss s Law states that the electic flu,, though a closed suface is given by S E da 4 kq in q in / o Results of the application of Gauss s Law follow.
5 5 INSULATORS 1. Solid Sphee a > a E = k / 2 < a E = k / a 3 > 0 adially outwad < 0 adially inwad 2. Spheical Shell a > a E = k / 2 < a E = 0 > 0 adially outwad < 0 adially inwad 3. Long Rod = / L E = 2k / > 0 adially outwad < 0 adially inwad 4. Lage Plane = / A E = 2 k = / 2 o > 0 nomal, outwad < 0 nomal, inwad 5. Two Lage Planes with Equal & Opposite Chage + between planes: E = 4 k = / o E - elsewhee: E = 0 nomal, fom + to - = / A *Note: unifom electic field between planes
6 6 CONDUCTORS IN ELECTROSTATIC EUILIBRIUM 1. Electic field inside the conducto is zeo. 2. Net chage inside the conducto is zeo. 3. Any net chage esides on the eteio suface(s) of the conducto. A metal spheical shell with a net chage of +90 nc and inne and oute adii of 4 cm and 5 cm is centeed aound a solid metal ball with a net chage of -50 nc and a adius of 2 cm. (a) Find the size and diection of the electic field at the following distances fom the cente of the solid ball. (i) 1.5 cm (ii) 3 cm (iii) 4.2 cm (iv) 6 cm (b) Find the amount of chage on the following sufaces. (i) outside suface of ball (ii) inside suface of shell (iii) outside suface of shell Ans. (a) (i) 0 (ii) 500,000 N/C, adially inwad (iii) 0 (iv) 100,000 N/C, adially outwad (b) (i) -50 nc (ii) +50 nc (iii) +40 nc
7 7 Motion of Chaged Paticles in E-Fields A chaged paticle of mass m moving in a egion of electic field epeiences an acceleation with a size of a = F m = q E m If the electic field is unifom in this egion then the acceleation is constant and the equations of motion fo constant acceleation can be used to descibe the paticle s motion. = o + v o t at2 v = v o + at v 2 = v o 2 + 2a( o ) = o (v o + v)t An electon, initially at est, is acceleated hoizontally fom left to ight by a unifom electic field of 2500 N/C between two metal plates that ae sepaated by 2 cm. Afte leaving these plates, the electon then entes a unifom field of 1000 N/C oiented vetically with the electic field pointing upwads. (a) Find the speed of the electon as it entes the second field egion. (b) Find the position of the electon 2 ns afte it entes the second field egion. Ans. (a) m/s (b) 8.4 mm to the ight and 0.35 mm down
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