two point charges of 2 c and -3c are separated by 6cm what is the magnitude of the electrostatic force between them? and is the force attractive or repulsive? explain and what happens to the magnitude of the force when the distance is increased by a factor of 2 to 12cm
Hint: Use Coulomb's Law.
Increasing the separation by a factor of 6 will decrease the (attractive) force by a factor of 36.
gegeg
To calculate the magnitude of the electrostatic force between two point charges, we can use Coulomb's Law. Coulomb's Law states that the magnitude of the electrostatic force (F) between two point charges (q1 and q2) separated by a distance (r) is given by:
F = (k * |q1 * q2|) / r^2
Where:
- F is the electrostatic force.
- k is the electrostatic constant, approximately equal to 9 × 10^9 N m^2/C^2.
- |q1| and |q2| are the magnitudes of the charges. Since the charges are given as 2C and -3C, we can take their absolute values.
Let's calculate the magnitude of the electrostatic force between the charges of 2C and -3C separated by a distance of 6cm.
|q1| = 2C
|q2| = |-3C| = 3C
r = 6cm = 0.06m
Plugging these values into Coulomb's Law, we get:
F = (9 × 10^9 N m^2/C^2 * |2C * 3C|) / (0.06m)^2
F = (9 × 10^9 N m^2/C^2 * 6C^2) / 0.0036m^2
F = (9 × 10^9 N m^2 * 6) / 0.0036
F ≈ 15 × 10^9 N
So, the magnitude of the electrostatic force between the charges is approximately 15 × 10^9 Newtons.
Now, let's determine if the force is attractive or repulsive. The charges involved are positive (2C) and negative (-3C). Since opposite charges attract each other, we can conclude that the force between these charges is attractive.
Next, you asked what happens to the magnitude of the force when the distance is increased by a factor of 2 to 12cm.
Using the same formula, with r = 12cm = 0.12m, we can calculate the new force:
F = (9 × 10^9 N m^2/C^2 * 6C^2) / (0.12m)^2
F = (9 × 10^9 N m^2 * 6) / 0.0144
F ≈ 37.5 × 10^9 N
So, when the distance is increased by a factor of 2 to 12cm, the magnitude of the force becomes approximately 37.5 × 10^9 Newtons.
To calculate the magnitude of the electrostatic force between two point charges, we can use Coulomb's Law, which states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * (|q1| * |q2|) / r^2
Where:
F is the magnitude of the electrostatic force
k is the Coulomb's constant (approximately 9 × 10^9 N m^2/C^2)
|q1| and |q2| are the absolute values of the charges
r is the distance between the charges
In this case, we have two point charges: 2C and -3C, separated by a distance of 6cm (0.06m). Let's calculate the magnitude of the force:
F = (9 × 10^9 N m^2/C^2) * ((2C * 3C) / (0.06m)^2)
Simplifying this equation, we get:
F = (9 × 10^9 N m^2/C^2) * (6C^2 / 0.0036m^2)
F = (9 × 10^9 N m^2/C^2) * 1666666.67 C / m^2
F ≈ 1.5 × 10^16 N
Therefore, the magnitude of the electrostatic force between the two charges is approximately 1.5 × 10^16 Newtons.
To determine whether the force is attractive or repulsive, we need to consider the signs of the charges. In this case, one charge is positive (2C) and the other is negative (-3C). Since opposite charges attract each other, the force between the charges is attractive.
Now, let's see what happens to the magnitude of the force when the distance is increased by a factor of 2 to 12cm (0.12m). We can plug this new distance into the equation and calculate the new magnitude of the force:
F = (9 × 10^9 N m^2/C^2) * ((2C * 3C) / (0.12m)^2)
F = (9 × 10^9 N m^2/C^2) * (6C^2 / 0.0144m^2)
F = (9 × 10^9 N m^2/C^2) * 416666.67 C / m^2
F ≈ 3.75 × 10^14 N
As we can see, when the distance between the charges is doubled, the magnitude of the electrostatic force decreases. In this case, the force decreases to approximately 3.75 × 10^14 Newtons.