3-01 Finite Line of Charge Due this Friday, Jan 19 at 11:59 pm (EST) A 14.5 cm l

Question

3-01 Finite Line of Charge Due this Friday, Jan 19 at 11:59 pm (EST) A 14.5 cm long rod with a uniform charge density of 2.0 m is shown below. What is the magnitude of the electric field at the point shown if d 40.0 cm? 8.25E4 N/C This question is based on an example worked in class, but you should practice setting up the integral because doing so is a possible exam question. I think it is easiest to set up if you define x-0 at the left end of the rod and use r=l+dx, but you can just plug the appropriate numbers into the equation I derived in dass using the center of the rod as the origin. Submit Answer Incorrect. Tries 6/15 Previous Tries

Explanation / Answer

According to the concept of the electrostatics

Given that

Charge density ¥=2*10^-6 c/m

Distance d=0.4 m

Length of the rod L=0.145 m

Now we find the electric field

Electric field E=K¥*L/d(L+d)

=9*10^9*2*10^-6*0.145/0.4(0.4+0.145)

=2.61*10^3/0.218

=11.97*10^3 N/c

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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