A positive charge Q1 = 30.5 nC is located at the origin. A negative charge Q2 = -3.5 nC is located on the positive x-axis p = 3 cm from the origin. calculate the location, x on the x-axis in centimeters where the electric field is equal to zero

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!!!

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so as we know that

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while tangential acceleration at the same moment will be zero

so

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