A line of positive charge is formed into a semicircle of radius R = 55.0 cm, as

Question

A line of positive charge is formed into a semicircle of radius R = 55.0 cm, as shown in Figure P23.63. The charge per unit length along the semicircle is described by the expression ? = ?0 cos ?. The total charge on the semicircle is 12.0 µC.

55.0 cm, as shown in Figure P23.63. The charge per unit length along the semicircle i A line of positive charge is formed into a semicircle of radius R described by the expression 3 no cos e. The total charge on the semicircle is 12.0 HC Figure P23.63 Calculate the total force on a charge of 2.00 uC placed at the center of curvature Magnitude Your response differs from the correct answer by more than 10%. Double check your calculations. NDirection upward O to the right downward O to the left

Explanation / Answer

The whole charge is the integral of _0 cos() R d from -pi/2 to pi/2. Therefore:
2 _0 R = 12.0 µC
So: _0 = (12.0e-6 C) / (1.1 m) = 10.909e-6 C/m

Because this static charge distribution is symmetric with respect to the vertical axis, the electrostatic field is vertical. Its value E = E_z at the center of curvature is given by the integral of the following integrand from -pi/2 to pi/2 (where the initial bracket [-cos ] account for the projection of the radial field on the vertical axis).
[-cos ] ( R d) / (4 pi epsilon_0 R^2)
= [ (cos )^2 d ] { -_0 / (4 pi epsilon_0 R) }
The integral of the square bracket from -pi/2 to pi/2 is equal to pi/2.
Therefore the integral of the whole thing is the (vertical) electrostatic field:
E_z = E = -_0 / (8 epsilon_0 R) = 280024.6935 N/C
For a (positive) charge of q = 2e-6 C, that field will produce a downward force:
q E = (2e-6 C) (280024.6935 N/C) = 0.56... N = 0.56 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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