Answer:
The points where the electric field due to the two charges are 0 include
x = 4.54 cm and x = 2.24 cm.
Explanation:
For a positive charge, the electric field is directed outward from the charge,
For a negative charge, it is directed towards it.
Since the positive charge is at the origin, the, and the negative charge is at p = 3cm on the x-axis, let the point where the electric field is 0 be x.
For ease of calculation, we will be looking for x, along the positive x-axis in between the two charges.
Electric field at a point due to a particular charge is given as
E = (kq/r²)
k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²
r = distance of the point from the charge.
q = the charge
q₁ = 30.5 nC = (30.5 × 10⁻⁹) C
q₂ = -3.5 nC = (3.5 × 10⁻⁹) C
For the positive charge
E₁ = (kq₁)/x²
= (9.0 × 10⁹ × 30.5 × 10⁻⁹)/x²
E₁ = (274.5/x²) (eqn 1)
For the negative charge
E₂ = [kq₂/(x-0.03)²]
E₂ = 31.5/(x-0.03)²
E₁ + E₂ = 0
(274.5/x²) - [31.5/(x-0.03)²] = 0 (with the assumption that the point is to the right of both charges)
(274.5/x²) =-[31.5/(x-0.03)²]
274.5 (x-0.03)² = 31.5x²
274.5 (x² - 0.06x + 0.0009) = 31.5x²
274.5x² - 16.47x + 0.24705 = 31.5x²
243x² - 16.47x + 0.24705 = 0
Solving this equation
x = 0.0454 m or x = 0.0224 m
So, the points where the electric field due to the two charges are 0 include
x = 4.54 cm and x = 2.24 cm.
Hope this Helps!!!
