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Oh I see, they are pointing in opposite directions in all these regions so the electric field ends up cancelling out.

Is it zero at the origin , z=0? But why would it be 0 at the origin , is it just a convention or is the strength of the electric field really 0 there ? The LaTeX doesn't appear in the message so I don't understand the points 1. 2. and 3.

I don't know why they are not opening , Hope its fixed now. Its png files.

TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2...

I'm doing this question to prepare for exams but i got stuck here. Because it's not in standard complex form, i don't know how to find angle theta. I could rationalize the denominator but i might end up with a messier expression.

no.

For a parallel RLC circuit, I have found the complex impedance to be 1/ (1/R -j(1/wL +wC)) . I need to find the phase difference between the voltage and current in the circuit. I know it's given by tan^-1(im(z)/re(z)) but how do I do it here as the expression is a fraction?

Hi, I have this pulley question, i have to find the force F that needs to be applied to the string to keep the system in equilibrium. I found 150N. Is it correct? T_1= 300N and T_2=150N so F is 150N.

I'm wondering if we had another string connected to m2 like I've shown in the diagram we would have to introduce another tension T3? And that tension would depend on the mass m2 only or It would be affected by the tensions T1 and T2 and we would get some other value? Because if I work out the...

But for a one pulley and 2 masses system in equilibrium, if the 2 masses are different, we still have the same tension? So that only applies to this example where the tensions are equal in magnitude if the masses 1 and 3 are equal.

sum of forces on m1 = ma so T1 - m1g = 0 T1 = m1g and T2=m2g.

The system is in equilibrium.

forces in y direction must be balanced i.e m2g= t1 cos a1 + t2 cos a2. the tensions are reversed in my FBD, t1 is on the left and t2 on the right.

Here's the figure with all the forces, t1 is same on either side of pulley(I'm treating it as if it was a 2 mass system so tension is the same on both sides) and t2 same on either side for the second pulley. That's what I did initially. I'm not sure if it's correct.

there are 3 forces acting on m2 , t1 t2 and it's weight. I'm not sure if t1 and t2 are equal or not. That's why i posted this question. I know for a single pulley system, the tension is the same on both sides but here we have 2 pullies and 3 masses.