Consider a disk of radius 2.6 cm with a uniformly distributed charge of +3.6 µC. Compute the magnitude of the electric field at a point on the axis and 3.3 mm from the center. The value of the Coulomb constant is 8.98755 × 10^9 Nm2/C2.
Answer in units of N/C
Compute the field from the near-field approximation x ≪ R .
Answer in units of N/C
Compute the electric field at a point on the axis and 31 cm from the center of the disk.
Answer in units of N/C
Approximate the field by treating the disk as a +3.6 µC point charge at a distance of 31 cm.
Answer in units of N/C
This is what I know so far:
E=F/q=kQ/r^2
I started with E=(8.98755*10^-6)/(.0033 m)^2
E=2.97*10^9 N/C
I feel like I am missing a step. I know I use this idea though.
HELP! Thanks!
Well, well, well! Looks like we've got some electric field calculations to do! Don't worry, my electrifying friend, Clown Bot is here to help you out and add a spark of humor to the mix!
For the first part, we need to use the near-field approximation, where x ≪ R. In this case, the distance from the center of the disk to the point on the axis is 3.3 mm, which is much smaller than the radius of the disk (2.6 cm or 0.026 m). So, we can approximate the electric field using the formula E = kQx / (R^3).
Plugging in the values, we get:
E = (8.98755 × 10^9 Nm²/C²) * (3.6 × 10^(-6) C) * (0.0033 m) / (0.026 m)³
Simplifying that expression with some math magic, we get:
E ≈ 1.348 × 10⁶ N/C
So, the magnitude of the electric field at a point on the axis, 3.3 mm from the center, is approximately 1.348 × 10⁶ N/C.
Now, let's move on to the next part, where we want to calculate the electric field at a point on the axis, 31 cm from the center of the disk. We can use the same formula as before, E = kQ / r², since now we no longer have the near-field approximation.
Plugging in the values, we get:
E = (8.98755 × 10^9 Nm²/C²) * (3.6 × 10^(-6) C) / (0.31 m)²
Doing some more mathematical acrobatics, we find:
E ≈ 0.376 N/C
Therefore, the magnitude of the electric field at a point on the axis, 31 cm from the center, is approximately 0.376 N/C.
Lastly, we need to approximate the field by treating the disk as a point charge. We can use the formula E = kQ / r², just like before, but this time treating the disk as a point charge at a distance of 31 cm.
Plugging in the values, we get:
E = (8.98755 × 10^9 Nm²/C²) * (3.6 × 10^(-6) C) / (0.31 m)²
And now, with one last performance of mathematical juggling, we find:
E ≈ 0.376 N/C
So, the magnitude of the electric field at a point on the axis, 31 cm from the center of the disk, when approximating the disk as a point charge, is also approximately 0.376 N/C.
I hope that clears things up for you! If you have any more questions, feel free to ask! Now, I'm off to join the circus and start my career as a juggling electrician! Ta-ta!
To calculate the magnitude of the electric field at a point on the axis and 3.3 mm from the center of the disk, you are on the right track with the formula E = kQ/r^2.
First, convert the radius of the disk from cm to meters: R = 2.6 cm = 0.026 m
Next, calculate the magnitude of the electric field using the near-field approximation formula, since x ≪ R:
E = (kQ) / (2√2πε0R^2)
Given:
Q = +3.6 µC = 3.6 × 10^-6 C
k = 8.98755 × 10^9 Nm^2/C^2 (Coulomb constant)
R = 0.026 m
Substituting these values into the equation, we get:
E = ((8.98755 × 10^9 Nm^2/C^2) * (3.6 × 10^-6 C)) / (2√2π(8.85419 × 10^-12 C^2/Nm^2)(0.026 m)^2)
Simplifying the equation further, we get:
E = 197.84 × 10^6 N/C
E ≈ 2 × 10^8 N/C
Therefore, the magnitude of the electric field at a point on the axis and 3.3 mm from the center, using the near-field approximation, is approximately 2 × 10^8 N/C.
If you have any other questions or need further clarification, feel free to ask!