Two particles are fixed on an x axis. Particle 1 of charge 47.9 μC is located at x = -5.87 cm; particle 2 of charge Q is located at x = 32.3 cm. Particle 3 of charge magnitude 52.8 μC is released from rest on the y axis at y = 15.1 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?
Give your answers in μC.
http://ca.answers.yahoo.com/question/index?qid=20110714100031AAk4Hfp
To find the value of Q, we can use Coulomb's Law, which states that the force between two charged particles is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's first calculate the distance between particle 2 and particle 3. Particle 2 is located at x = 32.3 cm, and particle 3 is at y = 15.1 cm. We can use the Pythagorean theorem to find the distance between them.
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the values:
Distance = √[(32.3 cm)^2 + (15.1 cm)^2]
Now we can calculate the force between particle 2 and particle 3 using Coulomb's Law:
F = k * (q1 * q2) / r^2
where k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
For part (a), since the initial acceleration of particle 3 is in the positive direction of the x axis, the force between particle 2 and particle 3 must also be in the positive x direction.
Using Coulomb's Law:
F = k * (Q * 52.8 μC) / (Distance)^2
Now we can equate this force to the force experienced by particle 3 in the x direction to find the value of Q.
F = m * a
Where m is the mass of particle 3 (which we can assume to be 1 kg for simplicity) and a is the initial acceleration.
Setting these two forces equal to each other:
k * (Q * 52.8 μC) / (Distance)^2 = m * a
Now we have an equation with one variable, Q. We can solve for Q by rearranging the equation:
Q = (m * a * (Distance)^2) / (k * 52.8 μC)
Repeat the above steps using the same method for part (b), where the initial acceleration of particle 3 is in the positive y direction.
To find the value of Q, we can use the principle of electrostatic forces and apply Coulomb's Law.
(a) If the initial acceleration of particle 3 is in the positive direction of the x-axis, that means there is a net force acting on particle 3 in the positive x-direction. This force is produced by the electric field created by particles 1 and 2.
The net force on particle 3 can be calculated as the sum of forces due to particle 1 and particle 2:
F_net = F_1 + F_2
The magnitude of the force between two charges can be expressed as:
F = k * |Q1 * Q2| / r^2
Where k is the electrostatic constant (k ≈ 8.99 * 10^9 N m^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
The force due to particle 1 can be calculated as:
F_1 = k * |Q_1 * Q_3| / r_1^2
Where Q_1 is the charge of particle 1, Q_3 is the charge of particle 3, and r_1 is the distance between particle 1 and particle 3.
Similarly, the force due to particle 2 can be calculated as:
F_2 = k * |Q_2 * Q_3| / r_2^2
Where Q_2 is the unknown charge of particle 2, Q_3 is the charge of particle 3, and r_2 is the distance between particle 2 and particle 3.
Since particle 3 is released from rest, the net force on it equates to the mass of particle 3 multiplied by its acceleration:
F_net = m * a
As the mass of particle 3 is not given, we can assume its mass to be m. Therefore:
m * a = k * |Q_1 * Q_3| / r_1^2 + k * |Q_2 * Q_3| / r_2^2
Plugging in the given values:
Q_1 = 47.9 μC = 47.9 * 10^-6 C
Q_3 = 52.8 μC = 52.8 * 10^-6 C
r_1 = 0.0587 m
r_2 = 0.323 m
We can rearrange the equation and solve for Q_2:
Q_2 = (m * a * r_1^2 * r_2^2) / (k * Q_3 * (r_1^2 + r_2^2))
(b) If the initial acceleration of particle 3 is in the positive direction of the y-axis, the force acting on particle 3 in the y-direction is due to the electric field created by the particles 1 and 2, but since particle 3 starts from rest, the force in the y-direction is 0.
Therefore, we can set up the equation:
F_y = F_1y + F_2y = 0
The y-component of the force due to particle 1 is:
F_1y = k * |Q_1 * Q_3| * sin(theta_1) / r_1^2
And the y-component of the force due to particle 2 is:
F_2y = k * |Q_2 * Q_3| * sin(theta_2) / r_2^2
Since particle 3 is on the y-axis, theta_1 and theta_2 will be 90 degrees. Therefore, sin(theta_1) and sin(theta_2) will equal 1.
So the equation becomes:
F_1y + F_2y = k * |Q_1 * Q_3| / r_1^2 + k * |Q_2 * Q_3| / r_2^2 = 0
We can solve for Q_2 using the same expression as in the x-direction:
Q_2 = -(m * a * r_1^2 * r_2^2) / (k * Q_3 * (r_1^2 + r_2^2))
Remember, Q_2 is negative in this case because the force acting on particle 3 needs to be in the negative y-direction.
Now, let's plug in the given values:
Q_1 = 47.9 μC = 47.9 * 10^-6 C
Q_3 = 52.8 μC = 52.8 * 10^-6 C
r_1 = 0.0587 m
r_2 = 0.323 m
Solving for Q_2 in both cases will give us the value of Q in μC.