To determine at which point the electric field vector is zero, we need to find the point where the electric field vectors due to both charges cancel each other out.
Let's denote the distance from the negative charge with charge -1.50 C to the point where the electric field is zero as "x". The distance from the positive charge with charge +6.85 C to the same point is given as 2 m - x (since they are 2 m apart).
According to Coulomb's Law, the electric field produced by a point charge is given by:
E = k * (|q| / r^2)
where E is the electric field magnitude, k is Coulomb's constant (8.99 x 10^9 N⋅m^2/C^2), |q| is the magnitude of the charge, and r is the distance from the charge.
The electric field due to the negative charge at point x is given by:
E_negative = k * (|-1.50 C| / x^2)
The electric field due to the positive charge at point 2 - x is given by:
E_positive = k * (|6.85 C| / (2 - x)^2)
Since we want the total electric field to be zero, we need to set the magnitudes of the two electric fields equal to each other:
E_negative = E_positive
k * (|-1.50 C| / x^2) = k * (|6.85 C| / (2 - x)^2)
|-1.50 C| / x^2 = |6.85 C| / (2 - x)^2
Simplifying this equation:
(1.50 C) / x^2 = (6.85 C) / (2 - x)^2
Cross-multiplying:
1.50 C * (2 - x)^2 = 6.85 C * x^2
Expanding:
1.5 C * (4 - 4x + x^2) = 6.85 C * x^2
6 - 6x + 1.5 x^2 = 6.85 x^2
Combine like terms:
6 + 0 x - 6.85 x^2 + 1.5 x^2 = 0
Simplifying:
-5.35 x^2 + 6 = 0
Rearranging:
5.35 x^2 = 6
Dividing both sides by 5.35:
x^2 = 6 / 5.35
x^2 ≈ 1.1215
Taking the square root of both sides:
x ≈ ± 1.06
The electric field vector will be zero at two points, approximately 1.06 m away from the negative charge and 2 - 1.06 = 0.94 m away from the positive charge.
Two point charges are 2.00 m apart as shown on the right. The charge of the first object is -1.50 C, while the second object has a charge of +6.85 C. At which point will the electric field vector be zero?
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