Math Problem Statement

Q5. A uniform line charge of (10 + 0.75๐‘…๐‘)ยต๐ถ/๐‘š is located on the z-axis. Find E in rectangular coordinates at ๐‘ƒ(3 + 0.3๐‘…๐‘, 4 โˆ’ 0.4๐‘…๐‘, 8 + 0.8๐‘…๐‘) if the charge exists from. (b) โˆ’โˆž โ‰ค ๐‘ง โ‰ค (8 + 0.8๐‘…๐‘) (Take RN=11)

Solution

To solve this problem, we need to calculate the electric field E\mathbf{E} at a point P(3+0.3RN,4โˆ’0.4RN,8+0.8RN)P(3 + 0.3\text{RN}, 4 - 0.4\text{RN}, 8 + 0.8\text{RN}) due to a uniform line charge density ฮป=(10+0.75RN)โ€‰ฮผC/m\lambda = (10 + 0.75\text{RN}) \, \mu\text{C/m} distributed along the zz-axis from z=โˆ’โˆžz = -\infty to z=(8+0.8RN)z = (8 + 0.8\text{RN}).

Given:

  • ฮป=(10+0.75RN)โ€‰ฮผC/m\lambda = (10 + 0.75\text{RN}) \, \mu\text{C/m}
  • P(3+0.3RN,4โˆ’0.4RN,8+0.8RN)P(3 + 0.3\text{RN}, 4 - 0.4\text{RN}, 8 + 0.8\text{RN})
  • Charge extends along the zz-axis from z=โˆ’โˆžz = -\infty to z=(8+0.8RN)z = (8 + 0.8\text{RN})
  • RN=11\text{RN} = 11

Step 1: Substitute RN value

First, substitute RN=11\text{RN} = 11 into the expressions for ฮป\lambda and the coordinates of PP.

ฮป=10+0.75(11)=10+8.25=18.25โ€‰ฮผC/m\lambda = 10 + 0.75(11) = 10 + 8.25 = 18.25 \, \mu\text{C/m} P(3+0.3(11),4โˆ’0.4(11),8+0.8(11))=P(3+3.3,4โˆ’4.4,8+8.8)=P(6.3,โˆ’0.4,16.8)P(3 + 0.3(11), 4 - 0.4(11), 8 + 0.8(11)) = P(3 + 3.3, 4 - 4.4, 8 + 8.8) = P(6.3, -0.4, 16.8)

Step 2: Use the electric field formula for a line charge

The electric field E\mathbf{E} due to a line charge along the zz-axis at any point (x0,y0,z0)(x_0, y_0, z_0) is given by:

E=ฮป2ฯ€ฯต0โˆซโˆ’โˆžzmax(x0i^+y0j^+(z0โˆ’z)k^)[(x02+y02+(z0โˆ’z)2)]3/2โ€‰dz\mathbf{E} = \frac{\lambda}{2\pi\epsilon_0} \int_{-\infty}^{z_{max}} \frac{(x_0\hat{i} + y_0\hat{j} + (z_0 - z)\hat{k})}{[(x_0^2 + y_0^2 + (z_0 - z)^2)]^{3/2}} \, dz

where:

  • zmax=8+0.8RN=16.8z_{max} = 8 + 0.8 \text{RN} = 16.8
  • ฯต0=8.854ร—10โˆ’12โ€‰F/m\epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}

Step 3: Simplify the integral

Given the symmetry of the problem, the zz-component of the electric field will not be zero due to the finite length of the line charge. The integrals for the xx and yy components are more complicated but can be computed with a direct substitution and integration.

Step 4: Calculating each component

Substituting ฮป\lambda and integrating:

  1. ExE_x:

    Ex=ฮป2ฯ€ฯต0โ‹…x0โˆซโˆ’โˆž16.81[x02+y02+(z0โˆ’z)2]3/2โ€‰dzE_x = \frac{\lambda}{2\pi\epsilon_0} \cdot x_0 \int_{-\infty}^{16.8} \frac{1}{[x_0^2 + y_0^2 + (z_0 - z)^2]^{3/2}} \, dz

  2. EyE_y:

    Ey=ฮป2ฯ€ฯต0โ‹…y0โˆซโˆ’โˆž16.81[x02+y02+(z0โˆ’z)2]3/2โ€‰dzE_y = \frac{\lambda}{2\pi\epsilon_0} \cdot y_0 \int_{-\infty}^{16.8} \frac{1}{[x_0^2 + y_0^2 + (z_0 - z)^2]^{3/2}} \, dz

  3. EzE_z:

    Ez=ฮป2ฯ€ฯต0โˆซโˆ’โˆž16.8(z0โˆ’z)[x02+y02+(z0โˆ’z)2]3/2โ€‰dzE_z = \frac{\lambda}{2\pi\epsilon_0} \int_{-\infty}^{16.8} \frac{(z_0 - z)}{[x_0^2 + y_0^2 + (z_0 - z)^2]^{3/2}} \, dz

Step 5: Compute the Electric Field Components Numerically

Finally, you can numerically evaluate the above integrals for each component of the electric field.

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Math Problem Analysis

Mathematical Concepts

Electrostatics
Vector Calculus
Electric Field

Formulas

Electric field due to a line charge: E = (ฮป / 2ฯ€ฮตโ‚€) โˆซ from -โˆž to z_max [(x_0 * รฎ + y_0 * ฤต + (z_0 - z) * kฬ‚) / ((x_0ยฒ + y_0ยฒ + (z_0 - z)ยฒ)^(3/2))] dz
Line charge density: ฮป = 10 + 0.75RN (ยตC/m)

Theorems

Coulomb's Law
Gauss's Law in Electrostatics

Suitable Grade Level

Undergraduate - Physics/Electromagnetism