Two equal charges repel one another with a force of 4.0 x 10-4 N when they are 10 cm apart. If they are moved until the separation is 20 cm, the repulsive force will be
this is an inverse-square relationship
doubling the distance will quarter the force
Well, if two equal charges repel each other with a force of 4.0 x 10-4 N when they are 10 cm apart, I wonder if they ever go on awkward "distance" dates? But I digress...
Let's use a little physics to figure out the repulsive force when the separation is 20 cm. We know that the force between these charges is inversely proportional to the square of the distance between them. So if we double the distance, the force should decrease by a factor of 4.
4.0 x 10-4 N / 4 = 1.0 x 10-4 N
Therefore, the repulsive force will be 1.0 x 10-4 N when the separation is 20 cm. Just like a clown's shoe, it all makes sense now!
To find the repulsive force between the two charges when they are moved to a new separation distance, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is given by:
F = k * (q1 * q2) / r^2
where:
F is the force between the charges,
k is the electrostatic constant (9 x 10^9 N*m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
r is the separation distance between the charges.
Let's calculate the new force.
Given:
Initial separation distance, r1 = 10 cm = 0.10 m
Initial force, F1 = 4.0 x 10^-4 N
We can write the equation as:
F1 = k * (q1 * q2) / r1^2
Solving for q1 * q2, we get:
(q1 * q2) = (F1 * r1^2) / k
Now, let's find the new separation distance, r2 = 20 cm = 0.20 m
The new force, F2, can be calculated as:
F2 = k * (q1 * q2) / r2^2
Substituting the values, we have:
F2 = k * [(F1 * r1^2) / k] / r2^2
Simplifying:
F2 = F1 * (r1^2 / r2^2)
Let's plug in the values and calculate:
F2 = (4.0 x 10^-4 N) * (0.10 m)^2 / (0.20 m)^2
F2 = (4.0 x 10^-4 N) * 0.01 / 0.04
F2 = 0.00004 N / 0.04
F2 = 0.001 N
Therefore, the repulsive force between the two charges when they are moved until the separation is 20 cm is 0.001 N.
To calculate the repulsive force between two charges, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.
Coulomb's Law formula:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges
k is the electrostatic constant (9 x 10^9 N m^2/C^2)
q1 and q2 are the magnitudes of the charges
r is the separation distance between the charges
Given:
F1 = 4.0 x 10^-4 N (force when separation is 10 cm or 0.1 m)
r1 = 0.1 m (separation distance when the force is given)
r2 = 0.2 m (new separation distance)
First, we need to find the magnitude of the charges at the initial separation distance (q1 and q2) by rearranging Coulomb's Law:
F1 = k * (q1 * q2) / r1^2
Rearranging the formula to solve for q1:
q1 = (F1 * r1^2) / (k * q2)
Substituting the given values into the formula:
q1 = ((4.0 x 10^-4 N) * (0.1 m)^2) / ((9 x 10^9 N m^2/C^2) * q2)
Next, we need to find the force at the new separation distance (F2) using Coulomb's Law:
F2 = k * (q1 * q2) / r2^2
Substituting the values we found for q1:
F2 = (k * (q1 * q2) / r2^2
= (k * ((4.0 x 10^-4 N) * (0.1 m)^2) / ((9 x 10^9 N m^2/C^2) * q2) * q2) / (0.2 m)^2
Simplifying the formula:
F2 = (F1 * (r1 / r2)^2) / 4
Substituting the given values:
F2 = ((4.0 x 10^-4 N) * ((0.1 m) / (0.2 m))^2) / 4
Calculating the value:
F2 = ((4.0 x 10^-4 N) * 0.5^2) / 4
= ((4.0 x 10^-4 N) * 0.25) / 4
= (1.0 x 10^-4 N) / 4
= 2.5 x 10^-5 N
Therefore, when the separation distance between the two equal charges is increased to 20 cm, the repulsive force between them will be 2.5 x 10^-5 N.