In Figure 22-43, a nonconducting rod of length L = 5.00 cm has charge -q = -4.84

Question

In Figure 22-43, a nonconducting rod of length L = 5.00 cm has charge -q = -4.84 fC uniformly distributed along its length. Fig. 22-43 (a) What is the linear charge density of the rod? C/m (b) What is the magnitude of the electric field at point P, a distance a = 12.0 cm from the end of the rod? N/C (c) What is its direction? ? (counterclockwise from the positive x axis) (d) What is the electric field magnitude produced at distance a = 50 m by the rod? N/C (e) Repeat part (d) for a particle of charge -q = -4.84 fC that replaces the rod. N/C

Explanation / Answer

To find the Electric field at a pont on the x axis you have to take a small element dx of the rod say at a distance x from the left end of the rod, the electric field at a dist a due to this will be klambda dx/ (x+a)^2, to get the field due to the rod integrate from x=0 to x=L

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.