Calculating Electric Field Intensity at (0, 0, z)

AI Thread Summary
The discussion revolves around calculating the electric field intensity at the point (0, 0, z) due to a charge density distributed along a specific arc. The user outlines their approach using the formula for electric field intensity and integrates over the defined limits. They express uncertainty about their variable evaluations and the clarity of their notation. Feedback indicates that while the approach is correct, the notation is difficult to follow. The user is encouraged to clarify their steps for better understanding.
technicolour1
Messages
7
Reaction score
0

Homework Statement



Electric charge of density ρℓ = 5 μC/m is distributed along the arc r = 2 cm,
0 ≤ ∅ ≤ ∏/4, z = 0. Find the electric field intensity at (0, 0, z)

Homework Equations


dE=(dQ(r)/(4∏ε(r^2))) (unit vector R)

The Attempt at a Solution


Letting pL be the charge density, letting r^2 = mag(r)^3

pLdL = dQ
dL = rd∅
pLrd∅ = dQ, r=0.02

Note: Limits of integration are 0 -> ∏/4

E(r) = 5E-6*0.02/(4∏ε)∫(z*(unitvector z) - 0.02cos∅* (unit vector x) - 0.02sin∅*(unit vector y))/((z^2+0.0004)^(3/2))) d∅

=900∫(z*(unitvector z) - 0.02cos∅* (unit vector x) - 0.02sin∅*(unit vector y))/((z^2+0.0004)^(3/2))) d∅I found it kind of hard to use notation while typing, so feel free to ask me to clarify. I know what I just typed is extremely tangly, so thanks in advance for taking the time to look at it.

My question is, did I go about this problem the right way? I don't feel confident about the evaluation I did to get all variables in terms of ∅. Also, I brought the unit vector into the calculation so that the final answer's vector will depend on what's being plugged in. I figure since z must be a constant point charge, the integral can be easily evaluated once a z value is given.
 
Physics news on Phys.org
technicolour1 said:

Homework Statement



Electric charge of density ρℓ = 5 μC/m is distributed along the arc r = 2 cm,
0 ≤ ∅ ≤ ∏/4, z = 0. Find the electric field intensity at (0, 0, z)

Homework Equations


dE=(dQ(r)/(4∏ε(r^2))) (unit vector R)

The Attempt at a Solution


Letting pL be the charge density, letting r^2 = mag(r)^3

pLdL = dQ
dL = rd∅
pLrd∅ = dQ, r=0.02

Note: Limits of integration are 0 -> ∏/4

E(r) = 5E-6*0.02/(4∏ε)∫(z*(unitvector z) - 0.02cos∅* (unit vector x) - 0.02sin∅*(unit vector y))/((z^2+0.0004)^(3/2))) d∅

=900∫(z*(unitvector z) - 0.02cos∅* (unit vector x) - 0.02sin∅*(unit vector y))/((z^2+0.0004)^(3/2))) d∅I found it kind of hard to use notation while typing, so feel free to ask me to clarify. I know what I just typed is extremely tangly, so thanks in advance for taking the time to look at it.

My question is, did I go about this problem the right way? I don't feel confident about the evaluation I did to get all variables in terms of ∅. Also, I brought the unit vector into the calculation so that the final answer's vector will depend on what's being plugged in. I figure since z must be a constant point charge, the integral can be easily evaluated once a z value is given.

The result looks correct.

You're right. It is rather hard to follow.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top