Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of

Question

14 days ago

11

Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of

them.
(a)Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0g when released?
(b)Which way will they accelerate?

Physics

2 answers:

inna [2.2K] · Answer 1 · 2026-03-11T12:46:51+03:00

Answer:

a) n = 4.77 × 10^7

b) They will move apart from each other.

Explanation:

a) Newton's Second Law states that:

Force exerted by each sphere = (mass)(Acceleration)

Force exerted by each sphere = (8.55 × 10^-3)(25) = 0.21N

Let Q denote the charge of each sphere, therefore, from Coulomb’s Law:

F = K Q^2/d^2

Q^2 = (0.21)(15×10^-2)^2/(9×10^9)

Q^2 = 5.83 × 10^-23

Q = 7.63 × 10^-12 C

To determine how many electrons need to be added, the following equation can be used:

Q = ne

n = 7.63 × 10^-12/e

n = 4.77 × 10^7

b) Since both spheres are negatively charged (having electrons), they will repel each other and accelerate away.

ValentinkaMS [2.4K] · Answer 2 · 2026-03-11T12:47:46+03:00

Answer:

1.43 x 10¹⁷.

They will move away from each other.

Explanation:

The force acting on each charged sphere is determined as F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Assuming Q is the charge on each sphere

F = $\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}$

Using the values, 2.095 = $\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}$

We find that Q² = $\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}$

Thus, Q = 2.289 X 10⁻⁶

The quantity of electrons = Charge / charge of a single electron

= $\frac{2.289\times10^{-6}}{1.6\times10^{-19}}$

=1.43 x 10¹³.

They will accelerate away from each other.