Two point charges q1 and q2 are held 4.00 cm apart vertically. An electron released 3 cm from 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
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three point charges are aligned along the x axis .find the resultant force on q2 due to other two charge?
To solve this problem, we need to use Coulomb's Law and the principles of electrostatics.
Coulomb's Law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
First, let's determine the charges of q1 and q2.
Given:
- The acceleration of the electron is 8.95 × 10^18 m/s^2.
- The distance of the electron from the middle point is 3 cm.
Since the acceleration is directed upward, it means that the electron is being repelled by the charges q1 and q2, which also implies that q1 and q2 have the same sign since they are equidistant from the electron.
Let's denote the magnitude of both charges as q.
Using the equation of motion, F = ma, we can find the force experienced by the electron due to the charges q1 and q2.
m = mass of the electron = 9.11 × 10^-31 kg
a = acceleration of the electron = 8.95 × 10^18 m/s^2
The force experienced by the electron, F = ma
Let's calculate the force first:
F = (9.11 × 10^-31 kg) * (8.95 × 10^18 m/s^2)
F = 8.16 × 10^-12 N
Since the force is the same for both charges q1 and q2, we can write:
F = k * (q1 * qe) / r1^2 = k * (q2 * qe) / r2^2
Here, k is the electrostatic constant, qe is the charge of an electron, r1 is the distance between q1 and the electron, and r2 is the distance between q2 and the electron.
The electrostatic constant, k, is equal to 8.99 × 10^9 N m^2 / C^2, and the charge of an electron, qe, is equal to -1.6 × 10^-19 C.
Since the electron is at the midpoint between q1 and q2, we can write:
r1 = r2 = 2 cm = 0.02 m
Now, let's solve for the charges q1 and q2.
8.16 × 10^-12 N = (8.99 × 10^9 N m^2 / C^2) * ((q * -1.6 × 10^-19 C) * (q * -1.6 × 10^-19 C)) / (0.02 m)^2
Simplifying the equation:
8.16 × 10^-12 N = (8.99 × 10^9 N m^2 / C^2) * (2.56 × 10^-38 C^2) / (0.02 m)^2
Now, solve for q:
q = sqrt((8.16 × 10^-12 N) * (0.02 m)^2 / ((8.99 × 10^9 N m^2 / C^2) * (2.56 × 10^-38 C^2)))
q ≈ 5.78 × 10^-19 C
Since q1 and q2 have the same magnitude, they both have a charge of approximately 5.78 × 10^-19 C.
Therefore, the magnitude of q1 and q2 is approximately 5.78 × 10^-19 C.
For the direction, since the electron is being repelled by q1 and q2 in the upward direction, the charges q1 and q2 must be positive.
Hence, the direction of q1 and q2 is positive.