I don't know if it's possible to have circuits with a negative resistance ... I don't think so.
For part (b) I have to admit I'm no used to dealing with components in parallel, but as I recall the voltage across the R and C is the same but the currents through them are 90° out of phase. The total current has to be 5A, so you'd get 5 cos30 through the capacitor and 5 sin30 through the resistor.
So the resistance of R is 100/2.5 = 40Ω
And the reactance of C is Xc = 100/(2.5√3) = 40/√3Ω
I think with parallel circuits V is the same so you draw V on the horizontal axis. Then for a resistor I is horizontal, for an inductor I is vertically down and for a capacitor I is vertically up.
But I must admit I'm only used to phasors for series circuits.
But we can consider a few limits. Suppose the beam velocity v = 0, then to balance out the negative charge of n₀ electrons you need an equal number of positive charges i.e. n = n₀. Yes?
There are two reasons why I chose (a), neither entirely logical :-)
Firstly, if we consider increasing the velocity to v = c then (a) would give n = 0 while (b) would give n = 2n₀.
But n = 2n₀ doesn't seem special. Why a factor of 2? Why not 3 or 4? My feeling is that as v ⟶ c I would expect either n ⟶ 0 or n ⟶ ∞, not just to some factor like 2. So (b) doesn't seem likely for that reason.
Secondly, if we are considering some finite distance of travel then the faster the beam goes the less time it takes so the less time the electrons have to move apart from each other and therefore the less positive charge we would need to hold them together.
This also suggests n decreases as the beam moves faster.
But these are both kind of arm waving arguments - I can't think of a rigorous way to answer the question.
XL = 20 because we know the voltage across the inductor is 100V and the current through it is 5A. The reactance is then just |V| / |I| = 100/5 = 20Ω
On our phasor diagram we find Vrc is 30° below the horizontal axis, and on this diagram we have current horizontal. This is how we do the diagrams for series circuits.
To do parallel circuits we need the voltage along the horizontal axis, and we can do this by rotating the whole diagram 30° anticlockwise. This puts the voltage Vrc along the horizontal axis, and then the current is 30° above the horizontal axis.
To find the position for C, I instinctively equated the electrostatic force between A-C and B-C. That did give me the right answer, but I do not understand why.
By "clamped", do they mean that it would be fixed DUE TO the electrostatic force or fixed onto the table otherwise?
@Buraian oh, do you have the link for it? I couldn't find it anywhere.
Yes, it did give me the position. What I'm trying to ask is - why? I interpreted the question to mean that the particle would be clamped to the table "physically". Does it instead want to convey that the particle should get fixed where it is placed as a consequence of equal and opposite electrostatic attraction?
I understand now. I did that straight away when I saw the question instead of writing all that prior reasoning down. Probably I subconsciously remembered having solved a similar question somewhere else but now I understand how we reach there. Thank you!
For a small displacement of C perpendicular to the line joining A and B, what would be the necessary conditions for C to perform simple harmonic motion?
You also need the displacement to be small enough for the restoring force to be proportional to the displacement. At large displacements C will still oscillate but the motion is anharmonic.
The linear proportionality of the force comes from a binomial expansion doesn't it?
The question mentions the distance to be very small compared to d further on, so after making that approximation yes, the force on C does come out to be linearly proportional to the displacement.
Thank you! The question asked the necessary conditions (for SHM) but they weren't mentioned in the answer. Thought it'd be best to confirm.
Hello @JohnRennie sir, suppose we consider a cone and a cylinder and take two points $A$ and $B$ in it differing by a height $H$. Then what will be $|P_A-P_B|$ in both the cases?
Yes that is what I'm saying, to compare between the two, if we write Bernoulli's equation for both the vessels and rearrange then finally 1/2*rho*v^2 terms will be left
Hmm, after writing Bernoulli's equation for both the vessels, subtracting one of the equation from the other and rearranging the terms, I can see that option B implies that (v(b)^2 - v(d)^2 + v(c)^2 - v(a)^2) is positive
where v(i) = speed of water at point I
I don't think we can say what the sign of (v(b)^2 - v(d)^2 + v(c)^2 - v(a)^2) is without comparing the cross sections and using principle of continuity.
@JohnRennie sir when a q charge is inside a cubical surface then flux through cube is $\frac{q}{\epsilon_0}$, when it is outside it , then flux is zero
What will be the flux through a cubical surface if charge is on it ( for example at center of one of its face)
I have 2 mechanical objects stacked on one another, the bottom one is accelerating and the top one does not slide on the bottom one.
How do I make sense of the static friction on the object on top?
Only the bottom one is applied a force to accelerate.
*top one
To my current understanding: because the object on top does not slide on the object at bottom: F-F_f=0=ma. That makes no physical sense because the top one's acceleration to the floor is 0, implying only the bottom one moves and the upper one stays still.
Problem Solving Strategies
Problem Solving Strategies