Question 43

43. A potential difference V is maintained between two large, parallel conducting plates. An electron starts from rest on the surface of one plate and accelerates toward the other. Its speed as it reaches the second plate is proportional to 2

e 2-0 AU+AK- 2eV —qV+—mv — —ev + —m v

Question 51

12 Electric current Magnetic field at wire 2 from current in wire 1: 21tr Force on a length AL Of wire 2: F = 12ALB Force per unit length in terms Of the currents: Magnetic field F AL \_ go 1112

Forces between two parallel infinitely long current-carrying conductors: Magnetic Field on RS due to current in PQ is (in magnitude) 27T r Force acting on RS due to current 12 through it is 12 12 1 sin 90' or — 27T r 27T r B acts perpendicular and into the plane of the diagram by Right Hand Thumb Rule. So, the angle between I and Bl is 90 • . is length of the conductor. Magnetic Field on PQ due to current in RS is (in magnitude) 2TT r Force acting on PQ due to current l, through it is po 12 l, sin 90' or = 27 r 1,121 2TT r Force per unit length of the conductor is 27 r (The angle between I and B2 is 90 • and B2 Is emerging out) 2TT r

Question 53

53. A charged particle can move with constant velocity through a region containing both an electric field and a magnetic field only if the A (B) (D) (E) electric field is arallel to the ma etic field electric field is perpendicular to the magnetic field electric field is parallel to the velocity vector magnetic field is parallel to the velocity vector magnetic field is perpendicular to the velocity vector

Because the forces caused by the two fields must be in opposite directions in order to add to zero, and since the force of the magnetic field is perpendicular to the magnetic field and the force of the electric field is in the same line as that of the field, the two fields must be perpendicular.

Ê=qÊ

Question 58

+Q As shown in the figure above, six particles, each with charge +Q, are held fixed and are equally spaced around the circumference of a circle of radius R.

58. With the six particles held fixed, how much work would be required to bring a seventh panicle of charge + Q from very far away and place it at the center of the circle? (B) 41t€o 3 3 (D) 21t€o (E) zt€o R

The potential at the center of the circle is V = so the potential 41E0 R 1 6Q2 energy of a sixth charge will be U = so this equals the work 41teo R 27teo R required to bring it from "infinity" where the U = 0.

Question 65

65. A physics problem starts: "A solid sphere has 'It charge distributed uniformly throughout .. .' may be correctly concluded that the (A) electric field is zero everywhere inside the (B) (C) (D) (E) sphere electric field inside the sphere is the same as the electric field outside electric potential on the surface of the sphere is not constant electric potential in the center of the sphere is zero sphere is not made of metal

22-3 Applications of Gauss's Law Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius ro. Determine the electric field (a) outside the sphere (r > ro) and (b) inside the sphere (r < 1 1

Question 69

1 11 111 39. When a negatively charged rod is brought near, but does not touch, the initially uncharged electroscope shown above, the leaves spring apart (I). When the electroscope is then touched with a finger, the leaves collapse (Il). When next the finger and finally the rod are removed, the leaves spring apart a second time (Ill). The charge on the leaves is (B) C D (E) positive in both I and Ill negative in both I and Ill ositive in I ne ative in Ill ne ative in I sitive in Ill impossible to determine in either I or Ill

In I the electrons are driven onto the leaves. In Il the electrons are allowed to go to ground, so in Ill the leaves have a net positive charge.

Question 70

x x x PI x x x Copper x x x P2 x x x -e 'O. A sheet of copper in the plane of the page is connected to a battery as shown above, causing electrons to drift through the copper toward the bottom of the page. The copper sheet is in a magnetic field B directed into the page. PI and P2 are points at the edges of the strip. Which of the following statements is true? (A) PI is at a higher potential than P2. (B) P2 is at a higher potential than Pl. (C) PI and P2 arc at equal positive potential. (D) PI and P2 are at equal negative potential. (E) Current will cease to flow in the copper sheet.

Negative charge will shift to the left causing the right side to be at a higher potential, since potential is defined in terms of the positive charge.

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