Electric charge is transferred to an irregular metallic disk as shown in the figure. If $ \sigma_1 $, $ \sigma_2 $, $ \sigma_3 $, and $ \sigma_4 $ are charge densities at given points, then choose the correct answer from the options given below:
Electric charge is transferred to an irregular metallic disk as shown in the figure. If $ \sigma_1 $, $ \sigma_2 $, $ \sigma_3 $, and $ \sigma_4 $ are charge densities at given points, then choose the correct answer from the options given below:
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- D and E Only
- A and C Only
- A, B, and C Only
- B and C Only
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to use the concept of surface charge density on a conductor. The key principle here is that charge tends to accumulate on parts of a conductive surface that are more curved, i.e., where the radius of curvature is smaller. This means sharper points on a conductor will have higher charge density.
Consider the points on the irregular metallic disk:
- \(\sigma_1\) is at a sharp corner, which generally has a higher curvature.
- \(\sigma_2\) and \(\sigma_4\) are at relatively flatter regions.
- \(\sigma_3\) is also at a sharper point, similar to \(\sigma_1\).
Based on these observations:
- The charge density, \(\sigma_1\), is likely greater than both \(\sigma_2\) and \(\sigma_4\) because \(\sigma_1\) is at a more curved (sharper) point.
- The charge density, \(\sigma_3\), might also be high because it shares high curvature (acuity) with \(\sigma_1\).
- The charge densities at \(\sigma_2\) and \(\sigma_4\) are likely lower since the surface is flatter, meaning these points have a larger radius of curvature.
Therefore, comparing the options:
- Option A: \(\sigma_1 > \sigma_3; \sigma_2 = \sigma_4\) - Possible as \(\sigma_1\) is in a more sharply curved area.
- Option B: \(\sigma_1 > \sigma_2; \sigma_3 > \sigma_4\) - Possible as it maintains the relative positions based on curvature.
- Option C: \(\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4\) - This aligns well with our understanding, as sharper points have higher density than flatter ones.
- Option D and Option E are incorrect based on the principles of charge accumulation.
The correct answer is: A, B, and C Only.
Approach Solution -2
In this problem, we are dealing with charge distribution on an irregular metallic disk.
The charge density on the surface of a conductor is not uniform and depends on the geometry of the conductor and the position on the conductor.
For a metallic disk:
- The charge densities at the edge are generally higher due to the fact that charges tend to accumulate at points of sharp curvature, such as the corners of the disk.
- The charge densities at the flat portions of the disk, away from the edges, are generally lower.
Looking at the figure:
- \( \sigma_1 \), being near the top edge of the disk, would have a higher charge density than \( \sigma_3 \), which is closer to the center.
- \( \sigma_2 \), being near the edge, would also have a higher charge density than \( \sigma_4 \), which is farther away from the edge.
Thus: - \( \sigma_1>\sigma_3 \)
- \( \sigma_2 = \sigma_4 \) (due to symmetry of the disk)
Therefore, the correct options are A, B, and C, meaning \( \sigma_1>\sigma_3 \), and \( \sigma_2 = \sigma_4 \).
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