Two point charges q1 and q2 are held 4.00 cm apart. An electron released at a point that is equidistant from both charges undergoes an initial acceleration of 8.95 X 10^18 m/s2 directly upward,parallel to the line connecting q1 and q2
Two point charges q1 and q2 are held 4.00 cm apart vertically. An electron released at the middle point that is equidistant from both charges undergoes an initial acceleration of 8.95 X 10^18 m/s2 directly upward,parallel to the line connecting q1 and q2 .Find the magnitude and direction of q1 and q2
To find the charges of q1 and q2, we can use Newton's law of electrostatic force. The formula for electrostatic force between two point charges is:
F = k * (|q1| * |q2|) / r^2
where F is the electrostatic force, k is the electrostatic constant (9.0 × 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of charges q1 and q2, and r is the distance between the charges.
In this case, we know that an electron released at a point equidistant from both charges undergoes an acceleration of 8.95 × 10^18 m/s^2 directly upward, parallel to the line connecting q1 and q2. Since the electron has a negative charge, we can use the equation of force on a charged particle:
F = m * a
where F is the force, m is the mass of the particle, and a is the acceleration.
The mass of an electron is 9.11 × 10^-31 kg, so we can calculate the force exerted on the electron:
F = (9.11 × 10^-31 kg) * (8.95 × 10^18 m/s^2)
F = 8.16 × 10^-12 N
Since this force is a result of the electrostatic force between the charges q1 and q2, we can set it equal to the formula for electrostatic force:
8.16 × 10^-12 N = (9.0 × 10^9 Nm^2/C^2) * (|q1| * |q2|) / (0.04^2 m^2)
Simplifying the equation, we find:
|q1| * |q2| = (8.16 × 10^-12 N) * (0.04^2 m^2) / (9.0 × 10^9 Nm^2/C^2)
|q1| * |q2| ≈ 1.82 × 10^-24 C^2
To solve for the charges of q1 and q2 individually, we need to make assumptions about their values. Let's assume |q1| = |q2| = q.
Then, we can rewrite the equation as:
q^2 ≈ 1.82 × 10^-24 C^2
Taking the square root of both sides, we get:
q ≈ √(1.82 × 10^-24 C^2)
q ≈ 4.27 × 10^-12 C
Therefore, both q1 and q2 have charges of approximately 4.27 × 10^-12 C.
You have described the physical situation.
What is the question?
Is the starting point of the electron along the line between q1 and q2? There are meny possible equidistant positions
I don't see how the acceleration can bbe both "directly upward" and parallel to the line between q1 and q2, unless q1 and q2 lie along a vertical line.