# A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in Figure P23.41. The charge per unit length along the semicircle is given by the expression λ = λ 0 cos θ . The total charge on the semicircle is 12.0 μ C. Calculate the total total on a charge of 3.00 μ C placed at the center of curvature P . Figure P23.41

### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278

Chapter
Section

### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278
Chapter 23, Problem 41AP
Textbook Problem
1108 views

## A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in Figure P23.41. The charge per unit length along the semicircle is given by the expression λ = λ0 cos θ. The total charge on the semicircle is 12.0 μC. Calculate the total total on a charge of 3.00 μC placed at the center of curvature P.Figure P23.41

To determine

The net force on a charge placed at the center of the curvature.

### Explanation of Solution

The radius of the semicircle is 60.0cm, the expression for the charge per unit area is λ=λ0cosθ and the total charge on the semicircle is 12.0μC.

Write the expression for the component of electric field along x axis

dE=dEcosθ=(KdQR2)cosθ                                                                                      (I)

Here, dQ is a segment of charge on the semicircle, dE is electric field due to a segment of charge, θ is the angle between the line joining the point and the charge with the vertical, K is the coulomb’s constant and R is the radius of the semicircle.

Write the expression for segmental arc length

dl=Rdθ

Here, dl is the arc length of the segment.

Write the expression for the total charge on the semicircle

dQ=λdlQ=π2π2λdl

Here, Q is the net charge on the semicircle and λ is the linear charge density.

Substitute λ0cosθ for λ and Rdθ for dl in above equation.

Q=π2π2λ0cosθRdθ=[λ0Rsinθ]π2π2=λ0(2R)λ0=Q2R

Substitute λdl for dQ, λ0cosθ for λ and Rdθ for dl in equation (I).

dEy=(KλdlR2)cosθ=(Kλ0cosθRdθR2)cosθ=(Kλ0R)cos2θdθ

Integrate the above equation to find the total electric field

E=π2π2dE=π2π2(K

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