The sum of two point charges is +11μC. When they are 2.7cm apart, each experiences a force of 224 N. Find the charges given that the force is:
a) repulsive
b)attraction
If repulsive, both charges are +
kq(11-q)/.027^2=224
solve for q
You will get two solutions in that quadratic, the one with different a negative sign for q will be the attractive solution.
To find the charges of the point charges, we can use the formula for the electric force between two charges:
F = k * |Q1 * Q2| / r^2
Where:
- F is the electric force between the charges
- k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- Q1 and Q2 are the magnitudes of the charges
- r is the distance between the charges
Now let's solve for the charges.
a) When the force is repulsive:
Given: F = 224 N, r = 2.7 cm = 0.027 m
Plugging the values into the formula and solving for |Q1 * Q2|:
224 N = (9 x 10^9 Nm^2/C^2) * |Q1 * Q2| / (0.027 m)^2
Simplifying the equation:
|Q1 * Q2| = 224 N * (0.027 m)^2 / (9 x 10^9 Nm^2/C^2)
|Q1 * Q2| = 0.0216 C^2
Since the charges are repulsive, they have the same sign. So |Q1 * Q2| = Q1 * Q2.
Therefore, Q1 * Q2 = 0.0216 C^2.
b) When the force is attractive:
Given: F = 224 N, r = 2.7 cm = 0.027 m
Plugging the values into the formula and solving for |Q1 * Q2|:
224 N = (9 x 10^9 Nm^2/C^2) * |Q1 * Q2| / (0.027 m)^2
Simplifying the equation:
|Q1 * Q2| = 224 N * (0.027 m)^2 / (9 x 10^9 Nm^2/C^2)
|Q1 * Q2| = 0.0216 C^2
Since the charges are attractive, they have opposite signs. So |Q1 * Q2| = -(Q1 * Q2).
Therefore, Q1 * Q2 = -0.0216 C^2.
Now we have two equations to solve for Q1 and Q2 in both cases.