The object A is a most likely conductor because field lines are not penetrating it.
(061)
Since the charge strength is proportional to the number of field lines starting/ending to the charged particle hence
for , number of field lines are 11 while for , it is 4 hence
-ve sign because field line goes into the charge which shows that it is negatively charged.
(062) Potential is largest at point 3
(063) The field is directed from left to right and negatively charged particle always moves opposite to the field line hence it will accelerate towards left
(064) The field is smallest at point 1 because here the field density is minimum
(065) Both point will have same potential bec they are the part of same body.
4. point 3 T2 5. point 4 063 (part 4 of 6) 0.5 points A negatively...
021 (part 1 of 3) 10.0 points An electron has an initial velocity of 3.5 x 10 m/s in the r direction. It enters a uniform electric field E = (300 N/C) which is in the y direction. Find the acceleration of the electron. The fundamental charge is 1.602 x 10-19 C and the mass of the electron is 9.109 x 10-31 kg. Answer in units of m/s". 022 (part 2 of 3) 10.0 points How long does it take...
The diagram at right shows an electric field line diagram and four labeled points. 2. (6 pts) Rank the magnitude of the electric field strength at points 1 to4 in the figure. (Remember "equal" is a possible ranking.") Indicate if the electric field strength is zero at any of the points. (8 pts) Give a good physics explanation for your ranking. This should consist of three parts. For example, if your ranking in a. is 1>2>3>4, you would first explain...
Only part B please 2. 3/6 points Previous Answers MI3 18.3.P.038. Fields of an electron At a particular instant a proton is at the origin, moving with velocity < 5 x 10", 4 x 10^, .5 x 104 > m/s. At this instant: (a) What is the electric field at location < 2 x 10,3 x 10,3 x 103 > m, due to the proton? E-< 2.790-5 , 4.186e-5 , 4.186e-5 > N/C (b) What is the magnetic field at...
Question 6, chap 123, sect 7. part 1 of 2 10 points Three point charges are placed at the ver- tices of an equilateral triangle. 60° Find the magnitude of the electric field vec- tor Ell at F. kq 5 kq 2 kq 2 k q 4 h ..2k q 1 kq 4 kq 1 kq 10. IE Question 7, chap 123, sect 7. part 2 of 2 10 points Find the direction of the field vector E at 0...
009 (part 1 of 3) 10.0 points A proton travels with a speed of 4.8×106 m/s at an angle of 96◦ west of north. A magnetic field of 4.6 T points to the north. Find the magnitude of the magnetic force on the proton. (The magnetic force experienced by the proton in the magnetic field is proportional to the component of the proton’s velocity that is perpendicular to the magnetic field.) Answer in units of N. 010 (part 2 of...
002 (part 1 of 3) 10.0 points A proton has an initial velocity of 3.62 x 107 m/s in the horizontal direction. It enters a uniform electric field of 21500 N/C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0.056 m horizontally. The mass of the proton is 1.6726 x 10-27 kg and the fundamental charge is 1.602 x 10-19 C. Answer in units of ns. 003 (part 2 of 3) 10.0 points What...
4 5 2 2 0.5 2.5 1 2 3.5 6 1.5 2 1 6.5 7 3 3 3 1 1.54 6.5 3 2 4 1 2 2 2.5 4 7 1 5 1 2 3 5 7 1 4 6 1.5 2 2 2 2 2 0.5 2 5 4 2 1 2.5 6 2.5 1 4 4 Use the data set provided in the main page of the experience for this application. 1. Complete the full hypothesis testing procedure...
[1 point] A negatively charged particle has a velocity in the negative z-direction at point P. The magnetic force on the particle at this point is in the negative y-direction. Which of the following statements about the magnetic field at point P can be determined from this data? [Notation: "B"represents the magnitude of the component of the magnetic field in the i-tih direction.] 1.) What information can be deduced about B,? a. Br is negative. b. B is positive. c....
Just need part 2 and 5. 3rd attempt Part 1 Part 2 Part 3 Part 4 Part 5 Parto e Part 1 Correct. Benzene is the inital starting compound of this synthesis. ® Part 2 Did you work backwards and forwards? Did you write a single molecular formula in this box? What chemical reagent is needed to complete this EAS reaction? Part 3 Correct. Benzene is brominated to form bromobenzene in the first step of the reaction Part 4 Correct....
part 1 and 2^part 3-6^ 10 9 10 9 4.00 x 109 > m what is the electric field at location < 10 9 2.00 x 10-9 A π 4.00 pi minus particle, which has charge-e, s at location < 7.00 10.9 > m, due to the π-particle? 4.00 2.00 E= -Select- An antiproton (same mass as the proton, charge-e) is at the observation location. what is the force on the antiproton, due to the π Select-- Lithium nucleus affected...