Question

An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the gravitational force on it. how far below the first electron is the second? the value of coulomb's constant is 8.98755 × 109 n · m2 /c 2 and the acceleration of gravity is 9.81 m/s 2 .

Answer 1

To cancel the gravitational force, the second electron exerts an electrostatic force on the first electron. The distance between the two electrons is approximately 5.42 × 10^-12 meters.

Step-by-step explanation:

To cancel out the gravitational force on the first electron, the second electron must exert an electrostatic force on it. Since the gravitational force acts downward, the electrostatic force must also act downward. Therefore, the second electron is located directly below the first electron. To find the distance between them, we can equate the electrostatic force to the gravitational force:

Fe = Fg

From Coulomb's law, we know that the electrostatic force between two charges is given by:

Fe = k(Q1Q2/r^2)

where k is Coulomb's constant, Q1 and Q2 are the charges of the electrons, and r is the distance between them.

Additionally, the gravitational force is given by:

Fg = mg

where m is the mass of the electron and g is the acceleration due to gravity.

Setting Fe equal to Fg, we have:

k(Q1Q2/r^2) = mg

Since the charges of electrons are equal and opposite, Q1Q2 is equal to -e^2, where e is the elementary charge (-1.602 × 10^-19 C).

Plugging in the values for Coulomb's constant (k = 8.98755 × 10^9 N · m^2 /C^2), the mass of the electron (m = 9.10938356 × 10^-31 kg), and the acceleration due to gravity (g = 9.81 m/s^2), we can solve for r:

8.98755 × 10^9 (-e^2/r^2) = (9.10938356 × 10^-31)(9.81)

Solving for r, we find that the distance between the first and second electrons is approximately 5.42 × 10^-12 meters.

Answer 2

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) = 8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926) meter²

D² = 25.84 m²

Take the square root of each side:

D = 5.08 meters

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?