A test charge of +5.0µC experiences a force of 2.0x103 N[S] when placed at the midpoint of two
oppositely charge parallel plates. Assuming that the plates are electrically isolated and have a
distance of separation of 8.0mm, what will be the force experienced by a different charge of -
2.0µC, located 2.0mm from the negative plate
E the same the whole way across so it does not matter how far from the plate.
F = Q E
E = voltage difference / distance between plates
so
(2/5) the force on the first one
To calculate the force experienced by a different charge of -2.0µC located 2.0mm from the negative plate, we can use Coulomb's law.
Coulomb's law states that the force between two charges is given by the equation:
F = k * |Q1 * Q2| / r^2
where:
F is the force between the charges,
k is the Coulomb's constant (k = 9.0 x 10^9 N m^2/C^2),
|Q1| and |Q2| are the magnitudes (absolute values) of the charges, and
r is the distance between the charges.
In this case, we have:
|Q1| = 5.0µC = 5.0 x 10^-6 C (positive charge in the midpoint of the plates)
|Q2| = -2.0µC = -2.0 x 10^-6 C (negative charge located 2.0mm from the negative plate)
r = 2.0mm = 2.0 x 10^-3 m (distance between the negative charge and the negative plate)
Plugging in the values into the equation, we get:
F = (9.0 x 10^9 N m^2/C^2) * |5.0 x 10^-6 C * -2.0 x 10^-6 C| / (2.0 x 10^-3 m)^2
Let's calculate the force.
F = (9.0 x 10^9) * (5.0 x 10^-6) * (-2.0 x 10^-6) / (2.0 x 10^-3)^2
F = -0.045 N
Therefore, the force experienced by the different charge of -2.0µC, located 2.0mm from the negative plate, is -0.045 N [S]. The negative sign indicates that the force is attractive.
To find the force experienced by a different charge, we can use Coulomb's Law which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:
F = (k * q1 * q2) / r^2
Where:
F is the force between the charges
k is the electrostatic constant, approximately 9.0 x 10^9 N·m^2/C^2
q1 and q2 are the charges (in coulombs) of the two charges
r is the distance (in meters) between the charges
First, we need to determine the magnitude of the force experienced by the +5.0µC test charge at the midpoint of the plates. The given force is 2.0x10^3 N[S] (newtons in the South direction). Since the plates are oppositely charged, the force experienced by the test charge in the North direction will also be 2.0x10^3 N (newtons).
Using Coulomb's Law, we can rearrange the formula to solve for the magnitude of one charge when the force is known:
F = (k * q1 * q2) / r^2
2.0x10^3 N = (9.0x10^9 N·m^2/C^2 * 5.0x10^-6 C * -5.0x10^-6 C) / (8.0x10^-3 m)^2
Now we can solve for q1 using algebra:
q1 = (F * r^2) / (k * q2)
q1 = (2.0x10^3 N * (8.0x10^-3 m)^2) / (9.0x10^9 N·m^2/C^2 * 5.0x10^-6 C)
Calculating this gives:
q1 = -2.8133 x 10^-6 C
Now, to find the force experienced by a different charge (-2.0µC) located 2.0mm from the negative plate, we can use Coulomb's Law again. Let's call this charge q3:
F = (k * q1 * q3) / r^2
F = (9.0x10^9 N·m^2/C^2 * -2.8133 x 10^-6 C * -2.0 x 10^-6 C) / (2.0x10^-3 m)^2
Calculating this gives:
F ≈ 16.66 N
So, the force experienced by the different charge of -2.0µC, located 2.0mm from the negative plate, is approximately 16.66 N.