Two small balls of masses 3m and 2m and each having charges Q are connected by a string passing over a fixed pulley. Calculate the acceleration of the balls (in `m//sec^(2)`) if the whole assembly is located in a uniform electric field `E=(mg)/(2Q)` acting vertically downwards. Neglect any interaction between the balls. Take `g= 10 m//s^(2)`

Answer

To solve the problem, we need to analyze the forces acting on each mass and apply Newton's second law of motion. Let's break it down step by step.

### Step 1: Identify the forces acting on each mass

1. **For the mass \(3m\)**:
- Weight acting downwards: \(W_1 = 3mg\)
- Electric force acting downwards: \(F_{electric} = QE\)
- Tension in the string acting upwards: \(T\)

2. **For the mass \(2m\)**:
- Weight acting downwards: \(W_2 = 2mg\)
- Electric force acting downwards: \(F_{electric} = QE\)
- Tension in the string acting upwards: \(T\)

### Step 2: Write the equations of motion for each mass

1. **For mass \(3m\)** (moving downwards):
\[ F_{net} = W_1 + F_{electric} - T = 3ma \]
Substituting the forces:
\[ 3mg + QE - T = 3ma \]

2. **For mass \(2m\)** (moving upwards):
\[ F_{net} = T - W_2 - F_{electric} = 2ma \]
Substituting the forces:
\[ T - 2mg - QE = 2ma \]

### Step 3: Substitute the value of the electric field

Given that \(E = \frac{mg}{2Q}\), we can substitute this into our equations.

- For the electric force:
\[ QE = Q \left(\frac{mg}{2Q}\right) = \frac{mg}{2} \]

### Step 4: Rewrite the equations with the electric force substituted

1. **For mass \(3m\)**:
\[ 3mg + \frac{mg}{2} - T = 3ma \]
Simplifying:
\[ \frac{6mg}{2} + \frac{mg}{2} - T = 3ma \]
\[ \frac{7mg}{2} - T = 3ma \quad \text{(Equation 1)} \]

2. **For mass \(2m\)**:
\[ T - 2mg - \frac{mg}{2} = 2ma \]
Simplifying:
\[ T - \frac{4mg}{2} - \frac{mg}{2} = 2ma \]
\[ T - \frac{5mg}{2} = 2ma \quad \text{(Equation 2)} \]

### Step 5: Solve the equations simultaneously

From Equation 1:
\[ T = \frac{7mg}{2} - 3ma \]

Substituting \(T\) into Equation 2:
\[ \frac{7mg}{2} - 3ma - \frac{5mg}{2} = 2ma \]
Simplifying:
\[ \frac{2mg}{2} - 3ma = 2ma \]
\[ mg - 3ma = 2ma \]
\[ mg = 5ma \]
Dividing both sides by \(m\):
\[ g = 5a \]
Thus, the acceleration \(a\) is:
\[ a = \frac{g}{5} \]
Substituting \(g = 10 \, \text{m/s}^2\):
\[ a = \frac{10}{5} = 2 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the balls is \(2 \, \text{m/s}^2\).

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