The electric field at the point x=5.00 cm and y=0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x=10.0 cm and y=0, the electric field points in the positive x direction with a magnitude of 15.0 N/C. Assuming the field is produced by a single point charge, find (a) it's location and (b) the sign and magnitude of its charge
Well, E = kq/r^2
E(5)=10
E(10)=15
E(5)=10=kQ/(p+.05)^2
E(10)=15=kQ/(p+.1)^2
divide the first equation by the second:
2/3 = [(P+.1)/(P=.05)]^2
take the sqrt of both sides, solve for P
Then put P into either equation, solve for Q
To find the location and charge of the single point charge, we can use the formula for the electric field created by a point charge:
\[
E = \dfrac{k \cdot |q|}{r^2}
\]
where:
- E is the electric field magnitude,
- k is Coulomb's constant, approximately 8.99 × 10^9 Nm²/C²,
- |q| is the absolute value of the charge,
- and r is the distance from the point charge.
To solve this problem, we need to find two equations by using the given information:
Equation 1: \(\dfrac{k \cdot |q|}{R_1^2} = 10.0 \, \text{N/C} \)
Equation 2: \(\dfrac{k \cdot |q|}{R_2^2} = 15.0 \, \text{N/C} \)
where \( R_1 = 5.00 \, \text{cm} = 0.0500 \, \text{m} \) is the distance from the point charge at (x=5.00 cm, y=0) and
\( R_2 = 10.0 \, \text{cm} = 0.100 \, \text{m} \) is the distance from the point charge at (x=10.0 cm, y=0).
Now, let's solve for the location and charge of the single point charge step-by-step:
Step 1: Solve for |q| using Equation 1.
Rearrange Equation 1:
\(|q| = \dfrac{E \cdot R_1^2}{k}\)
Substitute the given values:
\(|q| = \dfrac{10.0 \, \text{N/C} \cdot (0.0500 \, \text{m})^2}{8.99 \times 10^9 \, \text{Nm²/C²}}\)
|q| ≈ \(0.279 \times 10^{-8} \, \text{C}\) (rounded to three significant figures)
Step 2: Solve for |q| using Equation 2.
Rearrange Equation 2:
\(|q| = \dfrac{E \cdot R_2^2}{k}\)
Substitute the given values:
\(|q| = \dfrac{15.0 \, \text{N/C} \cdot (0.100 \, \text{m})^2}{8.99 \times 10^9 \, \text{Nm²/C²}}\)
|q| ≈ \(1.67 \times 10^{-8} \, \text{C}\) (rounded to three significant figures)
Step 3: Determine the sign of the charge.
Since the electric field points in the positive x-direction, the charge must be positive.
Step 4: Solve for the location of the single point charge.
Since the charge is positive, the charge must be located somewhere along the x-axis between x=5.00 cm and x=10.0 cm.
Therefore, the location of the single point charge is within this range.
To find the location and charge of the point charge, we can use Coulomb's Law equation. Let's assume the point charge is located at coordinates (x₀, y₀).
Coulomb's Law equation:
E = k * q / r²
Where:
- E is the electric field magnitude,
- k is Coulomb's constant (8.99x10^9 Nm²/C²),
- q is the charge of the point charge,
- r is the distance between the point charge and the point where the electric field is being measured.
(a) Finding the location of the point charge:
From the information given, we have two values of electric field magnitudes at different points. Let's use these two points to calculate the distances to the point charge.
Distance at x=5.00 cm and y=0:
r₁ = √[(x - x₁)² + (y - y₁)²]
r₁ = √[(5.00 cm - x₀)² + (0 - y₀)²]
Distance at x=10.0 cm and y=0:
r₂ = √[(x - x₂)² + (y - y₂)²]
r₂ = √[(10.0 cm - x₀)² + (0 - y₀)²]
Since the electric field points in the positive x-direction, the x-coordinate of the point charge must be between 5.00 cm and 10.0 cm.
(b) Finding the sign and magnitude of the charge:
We can determine the sign by considering the direction of the electric field. The electric field is positive in the x-direction, which suggests that the point charge is positive if the field points away from it and negative if it points toward it.
Now we have two equations involving the charge (q) and the distances (r₁ and r₂) that we derived from Coulomb's Law. We can solve these two equations to find the values of q, x₀, and y₀.
Using the given electric field magnitudes (E₁ = 10.0 N/C and E₂ = 15.0 N/C) along with Coulomb's Law, we can substitute the values and solve for q, x₀, and y₀.
Let's use these equations to solve for the charge and location of the point charge.
Since the E-field is pointed along the +x axis at locationa along the x axis, the point charge that causes the field but also be on the x axis. The distance from the point charge to (5,0) must be sqrt (2/3) = 0.8165 times the distance from the point to (10,0), because the field is 2/3 as strong at (5,0) as it is at (10,0).
The only way you can make this happen is to have the point charge be positive and located to the left x = 0.