A 10-cm-long thin glass rod uniformly charged to 13.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 13.0 nC are placed side by side, 3.50 cm apart. What are the electric field strengths E1 to 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?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3

The electric field is a vector magnitude so it must be added as vectors, in this case the field is requested on the line that joins the two charges, so we can enhance the ordinary sum.

E = E₁ + E₂

The two fields go in the same direction, so they add up

The expression for the electric field by a point charge is

E = k q / r₂

Where k is the Coulomb constant with value 8.99 10⁹ N m² / C², q is the load and r the distance of the charge to the test point

Let's calculate the electric field of a rod at a point r

Since the charge is evenly distributed, let's define a linear density

λ = q / x = dq / dx

dq = λ dx

The expression for the electric field is

E = k ∫ dq / x²

E = k λ ∫ dx / x²

E = k λ (-1 / x)

Let's evaluate between the limits infer x = r to the upper limit x = r + l

E = k λ (1 / r - 1 / r + l) = k λ l / r (r + l)

q = λ l

E = k q / r (r + l)

Now we can add the fields of each rod

a) r = 1 cm from the glass rod and this point is r = 3.5 -1 = 2.5cm from the plastic rod

E = k q / 0.01 (0.01 + 0.1) + k q / 0.025 (0.025 + 0.1)

E = k q (1 / 0.0011 + 1 / 0.003125) = 8.99 10⁹ 13.0 10⁻⁹ (909.09 + 320)

E = 1.44 10⁵ N / C

b) r = 2.0 cm from glass rod, r2 = 3.5-2 = 1.5 cm from plastic rod

E = k q (1 / 0.02 (0.02 + 0.1) + 1 / 0.015 (0.015 + 0.1))

E = 8.99 10⁹ 13 10⁻⁹ (1 / 0.0024 + 1 / 0.001725) = 116.87 (416.67 + 579.71)

E = 1.16 10⁵ N / C

c) r1 = 3.0 cm, r2 = 3.5-3.0 = 0.5 cm from the plastic rod

E = kq (1 / 0.03 (0.03 + 0.1) + 1 / 0.05 (0.05 + 0.1))

E = 116.87 (1 / 0.0039 + 1 / 0.0075) = 116.87 (256.41 + 133.33)

E = 4.55 10⁴ N / C