Two particles, with identical positive charges and a separation of
2.60×10−2m
are released from rest. Immediately after the release, particle 1 has an acceleration
→
a
1
whose magnitude is
4.60×103m/s2
while particle 2 has an acceleration
→
a
2
whose magnitude is
8.50×103m/s2
Particle 1 has a mass of
6.00×10−6kg
Find (a) the charge on each particle and (b) the mass of particle 2.

(a)Therefore the charge of each particle is .

(b)Therefore the mass of particle 2 is 3.25×10⁻⁶ Kg.

Explanation:

Given that, two positive charges  particles are equal charge and the distance between them is 2.60×10⁻² m²

The acceleration of the first particle is 4.60×10³ m/s².

The acceleration of the second particle is 8.50×10³ m/s².

The mass of the first particle is 6.00×10⁻⁶kg.

The coulombs Law:

F= Force between the charges.

= The charges

r= Distance between the changes.

Here ,r = 2.60×10⁻² m Nm²/C²

Therefore

  =1.33×10¹³ ×q² N.

Again we know that,

Force = mass× acceleration

For the first particle m₁=6.00×10⁻⁶kg, a₁= 4.60×10³ m/s² and F =1.33×10¹³ ×q² N

∴1.33×10¹³ ×q² =6.00×10⁻⁶×4.60×10³

C

Therefore the charge of each particle is .

(b)

To find the mass of the second particle we use the following formula,

Force = mass× acceleration

Here F=m₁a₁, mass = m₂, acceleration= a₂=8.50×10³ m/s².

m₁a₁= m₂a₂

⇒(6.00×10⁻⁶)×( 4.60×10³)= m₂×(8.50×10³)

        =3.25×10⁻⁶ Kg.

Therefore the mass of particle 2 is 3.25×10⁻⁶ Kg.

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