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.

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach