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hjlf
3 years ago
13

A long, nonconducting cylinder (radius = 6.0 mm) has a nonuniform volume charge density given by αr2, where α = 6.2 mC/m5 and r

is the distance from the axis of the cylinder. What is the magnitude of the electric field at a point 2.0 mm from the axis?
Physics
1 answer:
erastovalidia [21]3 years ago
5 0

Answer: 2.80 N/C

Explanation: In order to calculate the electric firld inside the solid cylinder

non conductor we have to use the Gaussian law,

∫E.ds=Q inside/ε0

E*2πrL=ρ Volume of the Gaussian surface/ε0

E*2πrL= a*r^2 π* r^2* L/ε0

E=a*r^3/(2*ε0)

E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C

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we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force that is responsible for the vibration. So what forces act upon a pendulum bob? And what is the restoring force for a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum. The tension force results from the string pulling upon the bob of the pendulum. In our discussion, we will ignore the influence of air resistance - a third force that always opposes the motion of the bob as it swings to and fro. The air resistance force is relatively weak compared to the two dominant forces.

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that's what I know so far

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3 years ago
When the voltage across a steady resistance is doubled, the current?
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I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.</span>

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But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

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