Two charges repel each other with a force of 10-5newton when they are 20cm apart. what is the force on each of them are i 3cm apart and ii when they are 100cm
Well, well, well. Looks like we've got some electric charges misbehaving here. Let's see what we can do with these questions:
i) When the charges are 3cm apart, the force between them can be found using the inverse square law. Since the distance is smaller, the force will be larger. But how much larger, you ask? To find out, we can calculate the ratio of the distances and square it. So, 3cm divided by 20cm gives us 0.15. Now we square that, and we get 0.0225. Finally, we multiply this by the original force of 10-5 Newton, and voila! The force on each charge when they are 3cm apart is approximately 2.25 x 10-6 Newton.
ii) Now let's stretch things out a bit and set the charges 100cm apart. Again, we'll use our trusty inverse square law. This time, the distance ratio is 100cm divided by 20cm, which gives us 5. Square that, and we get 25. Multiply this by the original force, and the force on each charge when they are 100cm apart is approximately 2.5 x 10-4 Newton.
Just remember, kids, keep your charges in check, and if things get too electrifying, call the nearest superhero for help!
To solve this problem, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's consider the given information:
Force (F1) = 10^-5 N
Distance (r1) = 20 cm = 0.2 m
Distance (r2) = 3 cm = 0.03 m
Distance (r3) = 100 cm = 1 m
i. When the charges are 3 cm apart:
We need to find the force on each charge. Since the charges are the same (repelling each other), their magnitudes are equal.
Using Coulomb's Law, we can write:
F1 / r1^2 = F2 / r2^2
Solving for F2:
F2 = F1 * (r2 / r1)^2
Plugging in the given values:
F2 = (10^-5 N) * ((0.03 m / 0.2 m)^2)
F2 ≈ 1.35 x 10^-6 N
Therefore, the force on each charge when they are 3 cm apart is approximately 1.35 x 10^-6 N.
ii. When the charges are 100 cm apart:
Similarly, we need to find the force on each charge, considering the new distance.
Using Coulomb's Law again:
F1 / r1^2 = F3 / r3^2
Solving for F3:
F3 = F1 * (r3 / r1)^2
Plugging in the given values:
F3 = (10^-5 N) * ((1 m / 0.2 m)^2)
F3 = 2.5 x 10^-4 N
Therefore, the force on each charge when they are 100 cm apart is approximately 2.5 x 10^-4 N.
To determine the force between two charges at different distances, we need to use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their 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 (k = 9 × 10^9 N m^2/C^2)
- q1 and q2 are the magnitudes of the charges
- r is the distance between the charges
Let's calculate the force on each of the charges at different distances:
i) When the charges are 3 cm apart:
Given: F = 10^(-5) N (force), r = 20 cm
To find the force at a different distance, we can first calculate the force at the initial distance and then use the ratio of the distances to find the force at the new distance.
Using Coulomb's Law, we have:
10^(-5) = k * (q1 * q2) / (0.2)^2
Rearranging the equation to find the product of charges:
(q1 * q2) = (10^(-5) * (0.2)^2) / k
Substituting the values:
(q1 * q2) = (10^(-5) * 0.04) / (9 * 10^9)
(q1 * q2) = 4 * 10^(-7) / (9 * 10^9)
Now, to find the force at a distance of 3 cm, we can use the ratio of distances:
F' = F * (r^2 / r'^2)
F' = (10^(-5)) * (0.2^2 / 0.03^2)
F' = (10^(-5)) * (0.04 / 0.0009)
F' = (10^(-5)) * 44.44
F' = 4.44 * 10^(-4) N
So, the force on each of the charges at a distance of 3 cm would be approximately 4.44 * 10^(-4) N.
ii) When the charges are 100 cm apart:
Using the same approach, we can calculate the force at this new distance.
F' = F * (r^2 / r'^2)
F' = (10^(-5)) * (0.2^2 / 1^2)
F' = (10^(-5)) * (0.04 / 1)
F' = (10^(-5)) * 0.04
F' = 4 * 10^(-7) N
So, the force on each of the charges at a distance of 100 cm would be 4 * 10^(-7) N.
You typing makes we wonder i i i
force is inversley prop
10E-5/F=( 3^2/20^2)
F=400/9 E-5 N at 3 cm
one 100 cm,that is 5 times as far..
F=1E-5 /25=4E-7 at 100cm