Fundamentals of Physics Electric Field

Electrical Field Any charge · when · has another force 2 STEP an associated electric charge interacts w/ this field field , that charge will experience a 1) Field FORCE is created 2) Force applied is The electrical · ° given by E : ↓= * electric field R1Q1 = any charged object at /r a distance r the in from = field Single Charge a Q is field distance This on E : to its separating charges single charge of Q describes surroundings * Also describes the 9 placed ** the in the resulting the effect electric the of force applied one on another field (Q) charge charge field points away from positive charges and points towards negative charges Y ↑ ↑ -Er The direction + > e of the electric field is defined F ↑ 44d as the direction that the positive charge electric would & move if it placed was at that location its always directed in the direction that the positive pulled if placed in the charge would be pushed space Surrounding the change or # what is the electric field at a distance 1 9 of . X0-Om from electron ? E = (k) /Q) r z = 14 . 0 X10") (1 (1 9 . . 6 x 10 X10 -8) 4) = : = 3488914 668 . 4 . 0 X100 N electron towards an Q # 1 what . 2 the electric field is electron ? E * (k) /Q) = (4 0x10 = 2(r)2 = . 6x10 -14) 3 00 = . X10-2 X10-0) 2 ( 1 P . 4 0x108 E = 2 (()(Q) 4 ~2 = ) (1 the away from (12) (Q) 12r) = % . twice the distance at j. 1 = . . 0 X10 ° NIC 4 SAMPLEQ#14 4 . electric E = field (12) /Q) at (4 = . 0 distance X10") (1 1 S ↓ 2 . . x at 1 . X10 "om 5 6 X 10") 10" 6 4 = °)2 . from proton ? X10" N Direction # A of electric field forces and fields positive charge electron F is parallel to electric field : negative charge F * The direction defined as charge that Q of = move F is = if q it was in opposite direction to a field is - > positive placed at a force experienced by 9X10'm from another electron ? E (4 0x100) F ( 1 6 x 10 9)(4 0x104) . ° - = = *The . F - - . 6 4 . q * Also t & #2 = = an qc- 10 - 13 because opposites attract value is I because the electric field moves electron - x if it is electro . x E a & location what is the electric 1 the N the electric the direction that would >° - > - electron R towards the positive Charge Q #2 What . 2 is the electric force experienced by froma Cathio e E qE = (1 6 X(0 = . 1 = = 1 28 . X10- . its 1 . PX10 % X(04) 0 N X10- N 3 . - ") (0 proton if a Q #3 field, both magnitude what is the electrical at q 92 point - = , 1 . D the in 21MC figure to the due q2r E 3 40NC = charges . (4 = . 6 83 x +Y . 0 X 13 (30 . 14 X10 3 : , 0 = · q Ez m 1 - 24 = . 07 . and 10") / 10 - m D +X . q direction and . 1 - . G 2) x 10 0X10") (3 40 X106) . - EFX & 6 56 = . EFy E Ex si = = = = = (5 . (Etx)" 02X10")2 6 00 . + - (EEy) (- 3 . EEy = = tan" 33 2 . (Ey/Ex) 3 . - E, X 16 56X10") 02 X10'"NIC . 07 = E = 20) = O )1 x , X10"N X 10's) - . cos30 2 20 x 10") X10'"NuC E - - , y - Ezy 56X10") Sin (30) 16 3 28 X . 3) 14 x 10-6) (6 83X10 6)2 1O - X 10's . [Enet = ? . 10'"NIC , Q #4 results using 10 2) a) . - MC 50 1 0 UC . E = answer b) F = 10-") (6 3 00 X 10 ° . 00 X electric qE = I -1 0 X 10. - . find , = 10 50 X = previous Charge charge qE = the from . 6 . (6 . force for 00 X 10") 00 X100 N 10' ") N . Electrical field lines · The lines As the lines (vice Versa , field) · · · Tangent on as the the field get lines become to the field Arrows Show direction lines field placed are in specific orientation and magnitude of the field electric the field stronger the relative closer , farther apart , the weaker the electric . never cross point of · electrical an indicate the direction and spacing · on o > q on (physically an directions E of intersection had lines Start - line indicate impossible or of else E at that location it means a negative charge the field multiple directions) positive charge C - Ea K & (4) ) and end at at Electric fields through symmetry component L in charged plates and , the field the direction L plate of charge has only perpendicular to the plate near a 7 7 t L t L 7 -Fields add they go 7 + L the because same direction + L a > 7 -Fields > because they go < L 7 y cancel opposite direction - 7 >= 7 > L n * Overall the larges * E is between I electric field the electric field to uniform inside the plates , L 0 = outside all , negatively charged the plates and magnitudes plates direction are all constant Q# negatively charged electric The field a) What b) What a) = is the Fg = 0 = Eg direction charge Down b) - E plates the 10" NIC 0 X = to the defecting platesw/ between the projected is once initial it enters position Vot /Vo 7 5 . 3 x x 10 m/25 16 "s mus : O Q #2 have electron beam points particle to the and direction) to mass W electron of W = -1 6 X10" J . 4 . = linetic initial an what is the , of X10-31 energy of electric uniform these electrons one 11 W WEk = right stop : beam have in a Is cm 1 . 6x10'"] field (magnitude distance ? leg Fe(r) cosO (gE) (r)cos180 = > - E W : (q) (v) Cos E : 100 1 6 x 10 -" - . 2 6 7 X10 = /(1 . 6 x10" ) (0 1Sm) cosId0 . NIC . Q #3 if a charged drop one -1 charge of a) what . 4x10 C and long does E . . Charge distance b) + += = x10" 1 13 = 4 75 = x 10 -1 = E m/25 . - 2 0 X10 looking = = = a + [10 1 6 . . m 10" ) (4 4 02S = . X 10 - 6 . 75 x 10") N - a m/s F /m = S = P 02S x = 0 . . for 4 position rot = (1 4 X = a = Oy qE = F I ma 0 0005 . + Eat 2 + 2 = Voy 0x10") (2 velocity of leg > - . 2) a) D/S = X10-1 / get through the electric field drop from straight path ? (Oy =? ( 4x10-10 . . a keg 0 0005 = 1 13 of field ? "N/C . mass uniform it take to 2) How much is the mass region of a + acceleration is b) how enters . 0 x 10 X10'm down - %12 0 . 0 x 2 0 . = 10. /1 13 0 X104 10" X10-5s O . m/s X10 m/s2 - 10 2smrs