Question 41

  • N=m·kg·s-2

  • J=m2·kg·s-2

  • C=A·s

  • V=m2·kg·s-3A-1

  • T=kg·s-2A-1

Question 48

  • C:\\266298A5\\73477446-49B2-471B-AFDD-BCD03931DCDD\_files\\image615.png

    A copper wire in the shape of a circle of radius I m, lying in the
plane of the page, is immersed in a mag- netic field, B, that points
into the plane of the page. The strength of B varies with time, t,
according to the equa- tion B(t) = 2t(1 - t) where B is given in
teslas when t is measured in seconds. What is the magnitude of the
induced electric field in the wire at time t = 1 s?

    Apply Faraday's Law of Electromagnetic Induction: dt E(2Ttr) - —(2t-
2t2) = 7tr2 (2 —4t) Since r = 1 m, the value of E at t = IsisE= 1 NIC.

Question 53

  • Parallel

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Question 56

insulating support 2R

  • Calculating Capacitance

    1. Assume a charge of +Q and -Q on each conductor

    2. Find the electric field between the conductors (Gauss's Law)

      • C:\\266298A5\\73477446-49B2-471B-AFDD-BCD03931DCDD\_files\\image620.png
    3. C:\\266298A5\\73477446-49B2-471B-AFDD-BCD03931DCDD\_files\\image181.png

      • C:\\266298A5\\73477446-49B2-471B-AFDD-BCD03931DCDD\_files\\image621.png
    4. C:\\266298A5\\73477446-49B2-471B-AFDD-BCD03931DCDD\_files\\image182.png

      • C:\\266298A5\\73477446-49B2-471B-AFDD-BCD03931DCDD\_files\\image622.png

Question 57

  • Since the magnetic force is always perpendicular to the object’s velocity, it does zero work on any charged particle.

  • Zero work means zero change in kinetic energy, so the speed remains the same.

  • Remember: The magnetic force can only change the direction of a charged particle’s velocity, not its speed.

Question 61


![If a conducting sphere contains a charge of + q within an inner cavity, a charge of —q will move to the wall of the cavity to "guard" the inte- rior of the sphere from an electrostatic field, regardless of the size, shape, or location of the cavity. As a result, a charge of +q is left on the exterior of the sphere (and it will be uniform). So, at points outside the sphere, the sphere behaves as if this charge

  • q were concentrated at its center, so the electric field outside is simply kQ/r . Since points X and Y are at the same distance from the center of the sphere, the elec- tric field strength at Y will be the same as at X. ](./media/image624.png)

Question 66

area vacuum vacuum

3εηΑ c 3εοΑ 3Κεο,4 2Kd+d 3κεμι 3Κε„Α 3εο,4

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