Three point charges +q, +2q and Q are placed at the three vertices of an equilateral triangle. Find the value of charge Q (in terms of q), so that electric potential energy of the system is zero.

Answer

To find the value of charge \( Q \) (in terms of \( q \)) such that the electric potential energy of the system is zero, we can follow these steps:

### Step 1: Understand the Configuration
We have three point charges placed at the vertices of an equilateral triangle:
- Charge \( +q \) at vertex A
- Charge \( +2q \) at vertex B
- Charge \( Q \) at vertex C

### Step 2: Write the Expression for Total Electric Potential Energy
The total electric potential energy \( U \) of the system is the sum of the potential energies between each pair of charges:
\[ U = U_{AB} + U_{BC} + U_{CA} \]
Where:
- \( U_{AB} \) is the potential energy between charges at A and B
- \( U_{BC} \) is the potential energy between charges at B and C
- \( U_{CA} \) is the potential energy between charges at C and A

### Step 3: Calculate Each Potential Energy
Using the formula for potential energy between two point charges \( U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} \), we can write:
1. For \( U_{AB} \):
\[ U_{AB} = \frac{1}{4 \pi \epsilon_0} \frac{(+q)(+2q)}{R} = \frac{2q^2}{4 \pi \epsilon_0 R} \]
2. For \( U_{BC} \):
\[ U_{BC} = \frac{1}{4 \pi \epsilon_0} \frac{(+2q)(Q)}{R} = \frac{2qQ}{4 \pi \epsilon_0 R} \]
3. For \( U_{CA} \):
\[ U_{CA} = \frac{1}{4 \pi \epsilon_0} \frac{(+q)(Q)}{R} = \frac{qQ}{4 \pi \epsilon_0 R} \]

### Step 4: Combine the Potential Energies
Now, substituting these into the total potential energy expression:
\[ U = \frac{2q^2}{4 \pi \epsilon_0 R} + \frac{2qQ}{4 \pi \epsilon_0 R} + \frac{qQ}{4 \pi \epsilon_0 R} \]
Factoring out \( \frac{1}{4 \pi \epsilon_0 R} \):
\[ U = \frac{1}{4 \pi \epsilon_0 R} \left( 2q^2 + 2qQ + qQ \right) = \frac{1}{4 \pi \epsilon_0 R} \left( 2q^2 + 3qQ \right) \]

### Step 5: Set Total Potential Energy to Zero
To find \( Q \) such that the total potential energy \( U = 0 \):
\[ 2q^2 + 3qQ = 0 \]

### Step 6: Solve for \( Q \)
Rearranging gives:
\[ 3qQ = -2q^2 \]
Dividing both sides by \( q \) (assuming \( q \neq 0 \)):
\[ 3Q = -2q \]
Thus,
\[ Q = -\frac{2}{3} q \]

### Final Answer
The value of charge \( Q \) in terms of \( q \) is:
\[ Q = -\frac{2}{3} q \]