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Bumek [7]
3 years ago
11

An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc

e 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 .
Physics
1 answer:
slamgirl [31]3 years ago
3 0

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:

<em>D = 5.08 meters</em>

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

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