# how many excess electrons must be placed on each of two small metal spheres placed 3.00 cm apart in a vacuum if the force of repulsion betwen the spheres is to be 0.000000008 N?

how do i solve this problem??=(

Use Coulomb's Law:
F = k Q^2/R^2

The force is F = 8*10^-9 N and the separation is R = 3*10^-2 m . Each sphere has the same charge, Q

Look up what k is and solve for Q. For the number of electrons N, solve
Q = N e
where e is the electron charge in Coulombs. Look that up too.

## Ah, the wonders of electron dating! Let's juggle some numbers and find a solution that'll cause an electrifying amount of laughter!

First, let's try to find the charge of each sphere. We have:

F = k * Q^2 / R^2

Plugging in the values, we get:

8 * 10^-9 N = k * Q^2 / (3 * 10^-2 m)^2

Now, if I remember correctly, the constant k in Coulomb's Law is approximately 8.99 * 10^9 N * m^2 / C^2. So let's substitute that value in and simplify:

8 * 10^-9 N = (8.99 * 10^9 N * m^2 / C^2) * Q^2 / (3 * 10^-2 m)^2

Phew! Looks like all the numbers are getting jiggy with it. Let's keep going!

Now, we can simplify further:

8 * 10^-9 = 8.99 * 10^9 * Q^2 / 9 * 10^-4

To solve for Q^2, we'll need to do some more math magic:

Q^2 = (8 * 10^-9) * (9 * 10^-4) / (8.99 * 10^9)

And after performing that calculation, we find:

Q^2 ≈ 8.91 * 10^-14 C^2

Woohoo! We're getting close! Now, let's find the charge of each sphere by taking the square root:

Q ≈ √(8.91 * 10^-14 C^2)

Grab a calculator, my friend, and solve that equation to find the value of Q. Just remember to use some gloves to handle those electrifying numbers!

## To solve this problem, you can follow these steps:

Step 1: Write down the known values:
- Force of repulsion (F) = 0.000000008 N (given)
- Separation between the spheres (R) = 0.03 m (given)

Step 2: Determine the value of the Coulomb constant (k):
- The Coulomb constant (k) can be looked up. Its value is approximately - 9 × 10^9 N m^2/C^2.

Step 3: Substitute the known values into Coulomb's Law:
F = k * (Q^2 / R^2)

0.000000008 N = (9 × 10^9 N m^2/C^2) * (Q^2 / (0.03 m)^2)

Step 4: Simplify and solve for Q^2:
0.000000008 N * (0.03 m)^2 = 9 × 10^9 N m^2/C^2 * Q^2

0.0000000000024 N m^2 = 9 × 10^9 N m^2/C^2 * Q^2

Step 5: Solve for Q^2:
Q^2 = (0.0000000000024 N m^2) / (9 × 10^9 N m^2/C^2)
Q^2 = 2.67 × 10^-22 C^2

Step 6: Take the square root of Q^2 to find Q:
Q = √(2.67 × 10^-22 C^2)
Q ≈ 5.17 × 10^-11 C

Step 7: Determine the number of excess electrons (N) on each sphere:
Q = N * e, where e is the charge of one electron

To find e, you can look it up as approximately 1.6 × 10^-19 C.

5.17 × 10^-11 C = N * (1.6 × 10^-19 C)

Step 8: Solve for N:
N = (5.17 × 10^-11 C) / (1.6 × 10^-19 C)
N ≈ 3.232 × 10^8

Therefore, approximately 3.232 × 10^8 excess electrons must be placed on each of the two small metal spheres for the force of repulsion between them to be 0.000000008 N.