Homework Statement
Homework Equations
F = I\vec{l} \times \vec{B}
\varepsilon = BLv
The Attempt at a Solution
Every direction i go seems real wrong
F = I\vec{l} \times \vec{B}
F = \frac{\varepsilon \vec{l} \times \vec{B}}{R}
F = \frac{\varepsilon LB}{R}
ma =...
ive always understood parallel/series as a visual type thing. youre saying 1/R + 1/3R = 4/3R -> 3R/4 ? I Have no idea what this does to the circuit, as in where this combined resistor would be.. relative to the middle one still standing
Homework Statement
Hi, this was an extra credit problem on a test I had that I think I remember correctly
Given R find R equivilent of the circuit
Homework Equations
Series R1, R2 = R1 + R2
parallel R1, R2 = (1/R1 + 1/R2)^-1The Attempt at a Solution
Never encountered a resistor in an...
Ok. thanks for all this btw.
Now here's a variation
For the switch being open, the capacitors on each side take all the voltage because i->0 after a long time. Va = V and Vb = 0, Vab = V
After the switch closes, there is a current path. This gives 2R some voltage, which means vb and va will...
The only thing I can conceptually be confused about is that you keep saying the caps will charge to that voltage source. but since the two right capacitors must add to V they can't... both, uh, be 1/3V. i think i misunderstand what you mean i guess
so when the switch closes and its been a long time the charge in the capacitors balance such that Vb = Va, and since nothing has changed on the left side, Va is the same making vb 1/3V?
right side
i know initially the charge on both capicitors are the same since theyre in series
q1=q2
and I know total V across is
V = \frac{q}{2c} + \frac{q}{c} = \frac{3q}{2c}
so
q = \frac{2CV}{3}
making V on capicitor1 1/3 V, capacitor 2 2/3V
Vb wrt ground would be 2/3 V since 1st...
Homework Statement
Given V R C for the figure, determine
a.
The voltage of a with respect to b when the switch is open?
b.
Which point a or b is at a the higher voltage?
c.
What is the final voltage of point b with respect to ground when the switch is closed?
d.
Wow much...
The concept of voltage being constant in a capacitor for some reason was lost on me.. that makes sense. this voltage is the same as the given correct? so
q = \frac{VL 2 \pi \epsilon_0}{\ln{1/3}}
C = \frac{L 2 \pi \epsilon_0}{\ln{1/3}}
oh, whoops. I meant ln(3R/2R) -> ln(3/2). you're saying use V(2R) as basis for the capacitance? I was kind of confused on how V(2R) contains all the charge we're talking about, but since the voltage I measured is from the surface of both conductor surfaces it should catch it all? I'm trying to...
Homework Statement
A cylindrical capacitor of length L consists of a solid conducting core with a radius R and an outer outer hollow conducting tube with an inner radius 3R. A voltage V_{ab} is applied between the two cylinders. Assume L >> R which means we can neglect edge effects.
Given [L...
y = \sqrt{\frac{3}{4}} a
ln term collapses to
ln{\frac{4}{3}}
Is this correct? What about part 3? I only know of 1/2mv^2 which I assume was at least greater than potential it would get out.. but I have no m given.
Homework Statement
Three rods each of uniform charge Q and length a are formed into an equilateral triangle. Given[Q,a,q] determine:
1. The voltage at the center of the triangle.
2. If a charge -q is placed at the center of the triangle determine potential energy.
3. Determine the initial...
Homework Statement
I got home from a test, had an extra credit problem semi memorized and am wondering how I was suppose to solve.
http://i.imgur.com/cIOdSvk.png
A plane with sides L is at x=3 . An electric field E = α y^2 x i + α z^2 y j passes through (where \alpha is a constant). Find the...