Question

A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 15.0 nC are placed side by side, 4.30 cm apart. What are the electric field strengths E1to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

Part A

Specify the electric field strength E1

Express your answer with the appropriate units.

Part B

Specify the electric field strength E2

Express your answer with the appropriate units.

Part C

Specify the electric field strength E3

Express your answer with the appropriate units.

Answer 1

a)

Q₁ = magnitude of charge on left rod = Q₂ = magnitude of charge on right rod = 15 x 10^-9 C

L = length of each rod = 10 cm = 0.1 m

r₁ = 1 cm = 0.01 m

r₂ = 3.30 m = 0.033 m

electric field at point 1 due to rod on left is given as

E₁ = k Q₁/(r₁ sqrt(r₁² + (L/2)²))

inserting the values

E₁ = (9 x 10⁹) (15 x 10^-9)/((0.01) sqrt((0.01)² + (0.1/2)²))

E₁ = 2.65 x 10⁵ N/C

electric field at point 1 due to rod on right is given as

E₂ = k Q₂/(r₂ sqrt(r₂² + (L/2)²))

inserting the values

E₂ = (9 x 10⁹) (15 x 10^-9)/((0.033) sqrt((0.033)² + (0.1/2)²))

E₂ = 0.68 x 10⁵ N/C

Net magnetic field at point 1 is given as

E = E₁ + E₂

E = (2.65 x 10⁵ ) + (0.68 x 10⁵ )

E = 3.33 x 10⁵ N/C

b)

Q₁ = magnitude of charge on left rod = Q₂ = magnitude of charge on right rod = 15 x 10^-9 C

L = length of each rod = 10 cm = 0.1 m

r₁ = 2 cm = 0.02 m

r₂ = 2.30 m = 0.023 m

electric field at point 2 due to rod on left is given as

E₁ = k Q₁/(r₁ sqrt(r₁² + (L/2)²))

inserting the values

E₁ = (9 x 10⁹) (15 x 10^-9)/((0.02) sqrt((0.02)² + (0.1/2)²))

E₁ = 1.25 x 10⁵ N/C

electric field at point 2 due to rod on right is given as

E₂ = k Q₂/(r₂ sqrt(r₂² + (L/2)²))

inserting the values

E₂ = (9 x 10⁹) (15 x 10^-9)/((0.023) sqrt((0.023)² + (0.1/2)²))

E₂ = 1.07 x 10⁵ N/C

Net magnetic field at point 2 is given as

E = E₁ + E₂

E = (1.25 x 10⁵ ) + (1.07 x 10⁵ )

E = 2.32 x 10⁵ N/C

c)

Q₁ = magnitude of charge on left rod = Q₂ = magnitude of charge on right rod = 15 x 10^-9 C

L = length of each rod = 10 cm = 0.1 m

r₁ = 3 cm = 0.03 m

r₂ = 1.30 m = 0.013 m

electric field at point 3 due to rod on left is given as

E₁ = k Q₁/(r₁ sqrt(r₁² + (L/2)²))

inserting the values

E₁ = (9 x 10⁹) (15 x 10^-9)/((0.03) sqrt((0.03)² + (0.1/2)²))

E₁ = 0.77 x 10⁵ N/C

electric field at point 3 due to rod on right is given as

E₂ = k Q₂/(r₂ sqrt(r₂² + (L/2)²))

inserting the values

E₂ = (9 x 10⁹) (15 x 10^-9)/((0.013) sqrt((0.013)² + (0.1/2)²))

E₂ = 2.01 x 10⁵ N/C

Net magnetic field at point 3 is given as

E = E₁ + E₂

E = (0.77 x 10⁵ ) + (2.01 x 10⁵ )

E = 2.78 x 10⁵ N/C