Recent content by Feeziks

Ok, let me try that again. So when r* > r *and* there is no other items around it, the internal sphere's E field is $$E = \frac{q}{ε4\pi r*^2} $$ This is the case where the external conducting sphere does not exist. However, the external hollow sphere exists, so E is not as above. I need to use...

" No, this is not a conductor. At radius s from the centre, consider separately the field due to the charge at radius less than s and that due to the charge at radius greater than s. " "You are overlooking this hint: you first need to find out how much charge is on the internal surface " "Using...

Ah, I thought that by dividing the E into E1 and E2 I had used that valuable information. I guess not. Let me try again. Also, for I thought that both the volume and the area utilises the same r? Since the internal sphere is fixed, so the volume and area should use the radius, r of the...

I am not quite sure how to present my answer in the form of a function with relation to the distance from the centre. What I got so far is the E1 and E2, for the internal and external sphere respectively. For internal sphere, the charge is volume * 𝜌, so it is $$ \frac{4\pi r^{3}}{3} * 𝜌$$...

I have a question, why is $$Q = 𝜆*𝜃$$? Since we know the sector length is S = R𝜃, so the length of the arc = S, and so I think it should be Q = 𝜆S = 𝜆R𝜃

I tried to substitute dE into the equation when solving for E, but I get stuck due to the limits and the d𝜃. If I integrate $$dE = \frac{k \lambda \, d\theta}{R}$$, if the limits are from 0 to infinity, I struggle to calculate due to the infinity involved...

I need to account for tension, weight, and repulsion. For the tension, I can draw the x and y component of Tmax and see that the x components of the 2 tensions Tmax will cancel out, and there are 2 y components of the Tmax to factor in. Weight is just F = mg, where g is acceleration due to...