a) Electric field strength at distances 1.0 cm from the glass rod along the line connecting the midpoints of the two rods E₁ = 5.04 x 10⁵ N / C
b) Electric field vector at distances 2.0 cm from the glass rod along the line connecting the midpoints of the two rods E₂ = 1.62 x 10⁵ N / C,
c) Electric field vector at distances 3.0 cm from the glass rod along the line connecting the midpoints of the two rods E₃= 1.75 x10⁵ N / C
<h3>What is Electric field?</h3>
The electric field is the addition of two electric field vectors. The electric field by a point charge is directly proportional to the charge and inversely proportional to the square of distance between them
E = E₁ + E₂ and E = k q / r
Where k is the Coulomb constant = 9 x10⁹ N m² / C², q is the charge and r the distance of the charge to the test point.
Suppose, calculate the electric field of a rod at a point r
Since the charge is evenly distributed, linear density will be
λ = q / x = dq / dx
dq = λ dx
The differential electric field vector is
dE = k ∫ dq / x²
dE = k λ ∫ dx / x²
On integrating between the limits infer x = r to the upper limit x = r + l
E = k λ (-1 / x)
E = k λ (1 / r - 1 / r + l) = k λ l / r (r + l)
E = k q / r (r + l) where q = λ l
a) r = 1 cm from the glass rod and this point is r = 4.5 -1 = 3.5cm from the plastic rod
E₁ = k q / 0.01 (0.01 + 0.1) + k q / 0.035 (0.035 + 0.1)
E₁ = 9 x10⁹ x5 x 10⁻⁹ x 1120.73
E₁ = 5.04 x 10⁵ N / C
b) r1 = 2.0 cm from glass rod, r2 = 4.5-2 = 2.5 cm from plastic rod
E₂ = k q (1 / 0.02 (0.02 + 0.1) + 1 / 0.025 (0.025 + 0.1))
E₂ = 9 x10⁹ x5 x 10⁻⁹x 3616.66
E₂ = 1.62 x 10⁵ N / C
c) r1 = 3.0 cm from glass rod, r2 = 4.5-3.0 = 1.5 cm from the plastic rod
E₃= kq (1 / 0.03 (0.03 + 0.1) + 1 / 0.05 (0.05 + 0.1))
E₃= 9 x10⁹ x5 x 10⁻⁹ x 389.744
E₃= 1.75 x10⁵ N / C
Thus, all the electric field strengths are find out at respective distances.
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