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A 10-cm-long thin glass rod uniformly charged to 10.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 10.0 nC are placed side by side, 4.50 cm apart. What are the electric field strengths E₁ to E₃ at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

A. Specify the electric field strength E₁
B. Specify the electric field strength E₂
C. Specify the electric field strength E₃

User Dseifert
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Final answer:

Without specific formulas for finite line charges, we cannot calculate the exact electric field strengths at the specified distances from a charged glass rod near a charged plastic rod. The electric field would be the vector sum of the fields caused by each rod, and the magnitude and the direction depend on the distance from the rods and the charges on them.

Step-by-step explanation:

The student has asked about the electric field strengths E₁, E₂, and E₃ at distances 1.0 cm, 2.0 cm, and 3.0 cm from a uniformly charged thin glass rod placed near a uniformly charged thin plastic rod. To solve this, the principle of superposition is used, where the electric field due to each rod at the point of interest is calculated and then the net field is found by vector addition. Unfortunately, without additional information on the charge distribution and without a specific formula for the electric field due to a finite line of charge, it is not possible to calculate the exact values for E₁, E₂, and E₃. If we were dealing with point charges or infinitely long charged rods, we could use Coulomb's law or the formula for an infinite line charge respectively, but these do not apply directly to finite rods. The distribution of the charge and the geometry of the setup could significantly affect the resulting field, so a more specific approach is needed that takes into account the finite size of the rods.

Still, some qualitative assessments can be made. The fields from the glass and plastic rods will be in opposite directions along the line connecting their midpoints due to their opposite charges. Moreover, the magnitude of the electric field will vary based on the distance from the rods. Closer to the rods (such as at 1.0 cm), the field will be stronger, and as the distance increases (2.0 cm and 3.0 cm), it will decrease.

User Surajeet Bharati
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