The electric field 2.0 x 10-2 m from a certain charge has a magnitude of 1.0 x 105 N/C. The magnitude of the electric field 1.0 x 10-2 m from the charge is

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The electric field 2.0 x 10-2 m from a certain charge has a magnitude of 1.0 x 105 N/C. The magnitude of the electric field 1.0 x 10-2 m from the charge is

BlackTom AI · Accepted Answer

To compare electric fields at different distances from a point charge, recall that E ∝ 1/r^2 for a given charge.

First, note the known values: at r1 = 2.0 × 10^-2 m, E1 = 1.0 × 10^5 N/C. We want E2 at r2 = 1.0 × 10^-2 m.

Compute the ratio using the inverse-square law: E2 / E1 = (r1^2) / (r2^2).
Plugging in the distances: r1^2 = (2.0 × 10^-2)^2 = 4.0 × 10^-4, and r2^2 = (1.0 × 10^-2)^2 = 1.0 × 10^-4.
Thus, E2 / E1 = (4.0 × 10^-4) / (1.0 × 10^-4) = 4.
Therefore, E2 = 4 × E1 = 4 × (1.0 × 10^5) = 4.0 × 10^5 N/C.

Option-by-option check:
- 2.5 × 10^4 N/C: This would imply a much smaller field at the closer distance, contradicting the 1/r^2 dependence; it would be more than an order of magnitude too small given the distance halving.
- 5.0 × 10^4 N/C: This is only five-tenths of the correct value, suggesting a misapplication of the scaling, perhaps using a linear instead of inverse-square relation or incorrect distance ratio.
- 2.0 × 10^5 N/C: This is exactly half the correct magnitude, which would correspond to a linear decrease with distance or a miscalculation of the squared distance ratio.
- 4.0 × 10^5 N/C: This matches the proper inverse-square scaling when the distance is halved, since squaring the distance ratio gives a factor of 4.

In summary, recognizing E ∝ 1/r^2 and applying the squared distance ratio correctly leads to E2 = 4.0 × 10^5 N/C, while the other options arise from common mistakes like misapplying the proportionality or forgetting the squaring of the distance ratio.

APPHYSC142-031 with labs Exam Review - Final

The electric field 2.0 x 10-2 m from a certain charge has a magnitude of 1.0 x 105 N/C. The magnitude of the electric field 1.0 x 10-2 m from the charge is

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