An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 5.3 m from the point charge. Determine the charge required for this to happen.

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An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 5.3 m from the point charge. Determine the charge required for this to happen.  The acceleration due to gravity is 9.8 m/s² and the Coulomb constant is 8.98755 x 10⁹ N.m²/C².  Answer in units of C.

the charge required for this to happen is 1.7415 × 10¹⁹ C

Explanation:

Given the data in the question;

we know that gravitational force on electro = M×g

mass of electron m = 9.10938356 × 10⁻³¹ kilograms

charge of electron e = 1.60217662 × 10⁻¹⁹ coulombs

given that distance r = 5.3m and the Coulomb constant = 8.98755 x 10⁹ N.m²/C²

g = 9.8 m/s²

Now for the electron to be suspended, This will happen only if the electrostatic force is equal to weight of electron;

[ (1/4πε0) × q × e] / r²   =   m × g

so we substitute

8.98755 x 10⁹ × q × 1.60217662 × 10⁻¹⁹/[5.3]² = 9.10938356 × 10⁻³¹ × 9.8

q =  8.9271 × 10⁻³⁰ / 5.1262 × 10⁻¹¹

q = 1.7415 × 10¹⁹ C

Therefore the charge required for this to happen is 1.7415 × 10¹⁹ C

[1.43996 × 10⁹ × q] / 29.09 =  8.92719 x 10⁻³⁰

1.43996 × 10⁹ × q  = 2.5969 × 10⁻²⁶

q = 2.5969 × 10⁻²⁶ / 1.43996 × 10⁹

q =  

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j