Recent content by palaphys

Agreed. This is simpler. Thanks

I think I will do it this way- for the entire wire covering the arc, the lengthwise turn density is N/(arc length). here arc length will be ## s= R\phi_{\max} = R\pi/3 ## based on the geometry of the problem. so ##dN/ds= N/s= \frac{3N}{\pi R}## where ##ds=R(d\theta) ## where theta is the polar...

I'm not able to edit it right now, but thanks for pointing it out. how is that possible? even their solution has included the effect of gravity right

I have tried to proceed with two methods, which agree with each other i) Energy analysis If we use work energy theorem, ## \sum W = \Delta K ## So, ## W_{\text{bat}} + W_{\text{field}} + W_{\text{grav}} = K_f \quad \text{(at } x = d/2 \text{)} ## Battery work: ## W_{\text{bat}} = V \cdot (C_i...

yes It must be proportional to some quantity. I'm trying to consider its relation with an angle theta, i.e polar angle

this is the figure provided. My approach was to consider the field due to an elemental ring-shape conductor, using the standard formula ##dB = \frac{\mu_0 i r^2 \ dN}{2(x^2 + r^2)^{3/2}} ## where r is the radius of the elemental ring, x is the distance from the center of the hemisphere to that...

I think I am satisfied with my solution, so I will continue using this kind of approach for such questions..

I agree that state 1 is the intended state. Also when I thought a bit more about this question, I think I understood it now, they are just saying that the energy of the bound K shell can be thought of as -27.5KeV in a way, so we can use ## E_L- E_K = E_{K_{alpha}} ## hence the "energy of the...

The question seems very simple, and its solution is also very simple. But what bothers me is the wording- "energy of Molybdenum atoms" what does that mean?? energy relative to what? also how can the energy of a bound system be positive? and I have never seen "energy of atoms" being used as a...

This post makes everything clear. Very high quality. All my doubts on this question got cleared! for a) I will take the direction of normal as rightward (as shown in diagram) b) ##\phi_{+q} = \frac{q}{10\varepsilon_0}## and ## \phi_{-q} = \frac{2q}{10\varepsilon_0}## c)...

1.##\frac{q}{\epsilon_0}## 2. ##\frac{-q}{\epsilon_0}## 3a)0 b)0

yes I wanted to do this

Intuition says zero.. flux due to one would "cancel out" the one due to the other

sorry radius of ring is given as 12cm. also for the flux calculation through the ring I ignored the charge on the ring, As I am unsure how to deal with self flux (if that exists) in electrostatics, and I am kind of confident that this question does not involve those complexities.(given its...

here is a simple diagram representing this situation. The issue I'm facing with this problem is a matter of sign convention. We can write the flux through the ring as the sum of the contribution of fluxes due to the point charges individually. So $$\Phi_{\text{net}} = \Phi_{+q} + \Phi_{-q} $$...