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

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SUMMARY

The discussion focuses on calculating the electric field intensity at the point (0, 0, z) due to a linear charge density of ρℓ = 5 μC/m distributed along an arc with radius r = 2 cm, spanning from 0 to π/4 radians. The user employs the equation dE = (dQ(r)/(4πε(r^2))) (unit vector R) and integrates to find the electric field, expressing the result in terms of the variable z. The evaluation of the integral is confirmed to be correct, although the notation used is noted as complex and potentially confusing.

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technicolour1
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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.
 
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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.
 

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