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

In summary: But I think the key part is the integral, and I think that's where you're going with this.In summary, the problem involves finding the electric field intensity at a point (0, 0, z) caused by a charge distribution of density ρℓ = 5 μC/m along an arc r = 2 cm, 0 ≤ ∅ ≤ ∏/4, z = 0. The approach involves using charge density pL and charge dQ to find the electric field intensity, and evaluating the integral with the appropriate limits of integration to get the final answer in terms of z. The use of unit vectors allows for the final answer to depend on the specific value of z. While the notation may be
  • #1
technicolour1
7
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.
 
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  • #2
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.
 

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

1. What is the formula for calculating electric field intensity at (0, 0, z)?

The formula for calculating electric field intensity at (0, 0, z) is E = kq/z^2, where E is the electric field intensity, k is the Coulomb's constant, q is the point charge, and z is the distance from the point charge to the point of interest.

2. How do you determine the direction of the electric field at (0, 0, z)?

The direction of the electric field at (0, 0, z) is determined by the direction of the electric field lines, which always point away from positive charges and towards negative charges. You can also use the right hand rule, where your fingers point in the direction of the electric field and your thumb points in the direction of the positive charge.

3. Can the electric field intensity at (0, 0, z) be negative?

Yes, the electric field intensity at (0, 0, z) can be negative if the point charge q is negative. This indicates that the electric field points in the opposite direction of the electric field lines.

4. What units are used to measure electric field intensity at (0, 0, z)?

The units used to measure electric field intensity at (0, 0, z) are Newtons per Coulomb (N/C) or Volts per meter (V/m).

5. How does the distance from the point charge affect the electric field intensity at (0, 0, z)?

The electric field intensity at (0, 0, z) is inversely proportional to the square of the distance from the point charge. This means that as the distance increases, the electric field intensity decreases. This relationship is described by the formula E = kq/z^2.

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