A thin glass rod is bent into a semicircle of radius R . A charge + Q is uniform

Question

A thin glass rod is bent into a semicircle of radius R . A charge +Q  is uniformly distributed along the upper half, and a charge ?Q  is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field E?  (in component form) at point P , the center of the semicircle in terms of k , Q , R , and ? . (Assume that point P  is located at the origin of a coordinate system where the positive x -direction is to the right and the positive y -direction is upwards.)

A thin glass rod is bent into a semicircle of radius R . A charge +Q is uniformly distributed along the upper half, and a charge ?Q is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field E? (in component form) at point P , the center of the semicircle in terms of k , Q , R , and ? . (Assume that point P is located at the origin of a coordinate system where the positive x -direction is to the right and the positive y -direction is upwards.)

Explanation / Answer

The E field from the top half is identical to the E field fromthe bottom half. Due to symmetry, only the vertical component ofeach bit of E field survives. . linear charge density = total charge / length ofrod = q / (? R / 2) = 2q/?R . Bit of charge: dq = ?ds . ds = R d? . dE = k dq / R2 . vertical component of dE is dEsin? where ? is from 0 to 90degrees . So now... total E field from top halfof rod is . integral of dE sin? = integral of sin? k (? Rd?) / R2 = . = k sin? (2q/?R) d? / R . = - ( 2kq / ?R2 )cos? evalfrom 0 to 90 you get . E = 2kq / ?R2 . This is only from the top half. The bottom half, due tosymmetry, contributes the same. So the total E field at P is twicethis value... . total E = 4 k q/?R2 straight downward note: k= 1/4??o The E field from the top half is identical to the E field fromthe bottom half. Due to symmetry, only the vertical component ofeach bit of E field survives. . linear charge density = total charge / length ofrod = q / (? R / 2) = 2q/?R . Bit of charge: dq = ?ds . ds = R d? . dE = k dq / R2 . vertical component of dE is dEsin? where ? is from 0 to 90degrees . So now... total E field from top halfof rod is . integral of dE sin? = integral of sin? k (? Rd?) / R2 = . = k sin? (2q/?R) d? / R . = - ( 2kq / ?R2 )cos? evalfrom 0 to 90 you get . E = 2kq / ?R2 . This is only from the top half. The bottom half, due tosymmetry, contributes the same. So the total E field at P is twicethis value... . total E = 4 k q/?R2 straight downward note: k= 1/4??o . This is only from the top half. The bottom half, due tosymmetry, contributes the same. So the total E field at P is twicethis value... . total E = 4 k q/?R2 straight downward note: k= 1/4??o

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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