Question

A strip of invisible tape 0.15 m long by 0.012 m wide is charged uniformly with a total net charge of 4 nC (nano = 1 times 10^-9) and is suspended horizontally, so it lies along the x axis, with its center at the origin, as shown in the figure below. Calculate the approximate electric field at location (0, 0.02, 0) m (location A) due to the strip of tape. Do this by dividing the strip into three equal sections, as shown in the figure below, and approx mating each section as a point charge. (Assume each point charge is seated at the center of the section it approximates. Express your answer in vector form.) What is the approximate electric field at A due to piece 1? What is the approximate electric field at A due to piece 2? What is the approximate electric field at A due to piece 3? What is the approximate net electric field at A? What could you do to improve the accuracy of your calculation? Use a smaller charge for piece 2 than for pieces 1 and 3: the actual electric field s similar to the field of a dipole, with most of the charge on the tape at the left and right ends. Use more sections: as the number of segments approaches infinity, the summation of the electric fields approximates the actual electric field. Use just two sections: due to symmetry about the y-axis, the components of the field due to each charge will be half of the total electric field. Use a larger charge for piece 2 than for pieces 1 and 3: since location A is closest to piece 2, this piece has the most significant contribution.

Answer 1

q = charge at the center of each sheet = 4 nC/3 = 1.33 x 10^-9 C

for sheet 1 on left :

$\theta$ = angle the line joining P to A makes with x-axis = tan^-1(0.02/0.05) = 21.8

r = distance P to A = sqrt(0.05² + 0.02²) = 0.054 m

Electric field is given as

E = k q/r² = (9 x 10⁹) (1.33 x 10^-9)/(0.054)² = 4.1 x 10³ N/C

b)

By piece 2 :

r' = 0.02 m

Electric field is given as

E' = k q/r'² = (9 x 10⁹) (1.33 x 10^-9)/(0.02)² = 3 x 10⁴ N/C

c)

By piece 3 :

$\theta$ = angle the line joining R to A makes with x-axis = tan^-1(0.02/0.05) = 21.8

r = distance P to A = sqrt(0.05² + 0.02²) = 0.054 m

Electric field is given as

E = k q/r² = (9 x 10⁹) (1.33 x 10^-9)/(0.054)² = 4.1 x 10³ N/C

d)

net electric field is given as

E_net = E Sin $\theta$ + E' + E Sin $\theta$

E_net = 2E Sin $\theta$ + E'

E_net = 2 (4.1 x 10³) Sin21.8 + (3 x 10⁴)

E_net = 3.3 x 10⁴

Answer 2

q = charge at the center of each sheet = 4 nC/3 = 1.33 x 10^-9 C

for sheet 1 on left :

$\theta$ = angle the line joining P to A makes with x-axis = tan^-1(0.02/0.05) = 21.8

r = distance P to A = sqrt(0.05² + 0.02²) = 0.054 m

Electric field is given as

E = k q/r² = (9 x 10⁹) (1.33 x 10^-9)/(0.054)² = 4.1 x 10³ N/C

b)

By piece 2 :

r' = 0.02 m

Electric field is given as

E' = k q/r'² = (9 x 10⁹) (1.33 x 10^-9)/(0.02)² = 3 x 10⁴ N/C

c)

By piece 3 :

$\theta$ = angle the line joining R to A makes with x-axis = tan^-1(0.02/0.05) = 21.8

r = distance P to A = sqrt(0.05² + 0.02²) = 0.054 m

Electric field is given as

E = k q/r² = (9 x 10⁹) (1.33 x 10^-9)/(0.054)² = 4.1 x 10³ N/C

d)

net electric field is given as

E_net = E Sin $\theta$ + E' + E Sin $\theta$

E_net = 2E Sin $\theta$ + E'

E_net = 2 (4.1 x 10³) Sin21.8 + (3 x 10⁴)

E_net = 3.3 x 10⁴