A positively charged particle is moving in the +y-direction when it enters a reg

Question

A positively charged particle is moving in the +y-direction when it enters a region with a uniform electric field pointing in the +x-direction. Which of the diagrams below shows its path while it is in the region where the electric field exists? The region with the field is the region between the plates bounding each figure. The field lines always point to the right. The x-direction is to the right; the y-direction is up. 6. A uniform linear charge of total charge 12.0 nC is distributed along the x axis from x = 0 to x = 3 m. What is the linear charge density (lambda) for this charge distribution? What is the electric field vector at y = 4 m on the y axis? Set up the Integrals only.

Explanation / Answer

6)  INTEGRAL[0,3] 72dx/(16+x^2)^3/2 is the answer because it has to be (9x10^9)(12x10^[-9])x4/(16+x^2)^3/2 in the negative x direction OR

ou have to sum up all the electric field contributions by point segments of the line of charge at (0,4).
One point segment has charge Q = dx
At point x, the vector direction toward the point (0,4) is (-x, 4). Therefore, the y component of the unit vector is 4 / sqrt(x^2 + 4^2).

Using the Coulomb's law,
E = k Q/r^2
here r^2 = x^2 + 4^2
And the question is asking for the y component of the electric field, so
Ey = k Q * 4 / (x^2+4^2) ^(3/2) = k dx* 4 / (x^2+4^2) ^(3/2)
If you integrate Ey with respect to x, you will have added up all the electric field contributions due to the line of charge.

= 4k x / 4^2 (x^2 + 4^2)^(1/2)
Now evaluate the expression above from x = 0 to x= 3
Ey = 1/4 * k [3 / 5] = 3/20 * k = 3/20 * 9e9 * 12e-9 = 16.2 N/C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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