When 2 point charges +q and +3q are held at a distance r from each other are released, they have an acceleration of a and 2a respectively. When we distribute the total charge equally between them , keep them at same distance as the original and release them , their accelerations now would be:
Initial Setup:
1. Charges and Forces:
- Two charges \( +q \) and \( +3q \) are separated by distance \( r \).
- The force between them initially is:
\[
F = \frac{k \cdot q \cdot 3q}{r^2} = \frac{3kq^2}{r^2}
\]
2. Accelerations:
- Let the masses of the charges be \( m_1 \) and \( m_2 \).
- Given that the initial accelerations are \( a \) and \( 2a \) for charges \( +q \) and \( +3q \) respectively, we have:
\[
F = m_1 a \quad \text{and} \quad F = m_2 \cdot 2a
\]
Step 1: Determine Mass Ratio
From the equations \( m_1 a = m_2 \cdot 2a \), we get:
\[
m_2 = \frac{m_1}{2}
\]
Step 2: New Charges After Redistribution
When the total charge \( q + 3q = 4q \) is equally distributed, each charge will be:
\[
\frac{4q}{2} = 2q
\]
Step 3: New Force Between Charges
With the new charges \( +2q \) and \( +2q \) at the same distance \( r \), the new force \( F' \) is:
\[
F' = \frac{k \cdot 2q \cdot 2q}{r^2} = \frac{4kq^2}{r^2}
\]
Step 4: New Accelerations
Using \( F' = m_1 a_1' \) and \( F' = m_2 a_2' \), we get:
1. For the first charge (\( m_1 \)):
\[
a_1' = \frac{F'}{m_1} = \frac{4kq^2 / r^2}{m_1} = \frac{4a}{3}
\]
2. For the second charge (\( m_2 = \frac{m_1}{2} \)):
\[
a_2' = \frac{F'}{m_2} = \frac{4kq^2 / r^2}{m_1 / 2} = \frac{8a}{3}
\]
Answer
The new accelerations are:
\[
\frac{4a}{3} \quad \text{and} \quad \frac{8a}{3}
\]
