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?
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.
To find the point at which the electric field vector is zero, we can use the principle of superposition. According to this principle, the net electric field at any point in space is the vector sum of the electric fields produced by the individual charges.
In this case, we have two point charges: one with a charge of -1.50 C and the other with a charge of +6.85 C.
Let's denote the distance of the unknown point from the -1.50 C charge as x, and the distance of the unknown point from the +6.85 C charge as (2.00 - x).
We can use Coulomb's law to find the electric fields produced by each charge at the unknown point. Coulomb's law states that the magnitude of the electric field produced by a point charge is given by:
E = k * |Q| / r^2
Where:
- E is the magnitude of the electric field
- k is Coulomb's constant (approximately 9.0 x 10^9 Nm^2/C^2)
- |Q| is the magnitude of the charge
- r is the distance between the charge and the point where the electric field is being measured
For the -1.50 C charge:
E1 = k * |-1.50| / x^2
For the +6.85 C charge:
E2 = k * |6.85| / (2.00 - x)^2
Since the electric field vectors produced by these two charges have opposite directions, the net electric field at the unknown point will be zero when the magnitudes of the electric fields are equal:
|E1| = |E2|
k * |-1.50| / x^2 = k * |6.85| / (2.00 - x)^2
To simplify, we can cancel out k:
|-1.50| / x^2 = |6.85| / (2.00 - x)^2
Now we can solve this equation for x.
|-1.50| * (2.00 - x)^2 = |6.85| * x^2
1.50 * (2.00 - x)^2 = 6.85 * x^2
Expand and rearrange the equation:
3.00 - 3.00x + x^2 = 6.85x^2
Combine like terms:
5.85x^2 + 3.00x - 3.00 = 0
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a
For our equation, the coefficients are:
a = 5.85
b = 3.00
c = -3.00
Substituting these values into the quadratic formula:
x = [-3.00 ± √(3.00^2 - 4(5.85)(-3.00))] / 2(5.85)
Simplifying:
x = [-3.00 ± √(9.00 + 70.20)] / 11.70
x = [-3.00 ± √(79.20)] / 11.70
x ≈ -0.139 m or x ≈ 1.401 m
Since we're looking for a point in between the two charges, the solution is x ≈ 1.401 m.
To determine the point at which the electric field vector is zero, we can use the principle of superposition. The electric field at a point is the vector sum of the electric fields created by each individual charge.
Let's assume that the point charges are labeled as Q1 (-1.50 C) and Q2 (+6.85 C). We want to find the position where the net electric field is zero.
To determine this, we can start by analyzing the electric field created by each charge separately.
The electric field created by a point charge can be calculated using the formula:
E = k * (Q / r^2)
Where E is the electric field, k is the electrostatic constant (9.0 x 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge to the point.
For Q1 (-1.50 C):
Let's denote the distance from Q1 to the point as x1.
The electric field created by Q1 at the point can be calculated as:
E₁ = k * (Q1 / x₁^2)
For Q2 (+6.85 C):
Let's denote the distance from Q2 to the point as x2.
The electric field created by Q2 at the point can be calculated as:
E₂ = k * (Q2 / x₂^2)
To find the point where the net electric field is zero, E₁ and E₂ need to cancel each other out. This means that their magnitudes should be equal:
E₁ = E₂
k * (Q1 / x₁^2) = k * (Q2 / x₂^2)
By rearranging the equation, we can solve for x2:
(Q2 / x₂^2) = (Q1 / x₁^2)
x₂^2 = (Q2 * x₁^2) / Q1
x₂ = sqrt((Q2 * x₁^2) / Q1)
Substituting the given values of the charges (Q1 = -1.50 C, Q2 = 6.85 C) and assuming x₁ = 2.00 m (as given in the question ), we can calculate the value for x₂.
x₂ = sqrt((6.85 C * (2.00 m)^2) / -1.50 C)
x₂ = sqrt((6.85 * 2^2) / -1.50) m
x₂ = sqrt(27.4 / -1.50) m
x₂ = sqrt(-18.27) m
Since the expression inside the square root is negative, which is not physically meaningful, it indicates that there is no point between the two charges where the net electric field is zero.