Two equal charges, +Q each, are at a distance r from each other. A third charge is placed on the line joining the above two charges such that all these charges are in equilibrium. Find q, in terms of Q.

Answer

To solve the problem of finding the charge \( q \) in terms of \( Q \) when two equal charges \( +Q \) are placed at a distance \( r \) apart and a third charge \( q \) is placed on the line joining them such that all three charges are in equilibrium, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges \( +Q \) located at points A and B, separated by a distance \( r \). We place a third charge \( q \) at a point C on the line joining A and B. Let the distance from charge A to charge C be \( x \). Thus, the distance from charge B to charge C will be \( r - x \).

### Step 2: Apply the Condition for Equilibrium
For the charge \( q \) to be in equilibrium, the net force acting on it due to the other two charges must be zero. This means that the force exerted on \( q \) by charge A must be equal in magnitude and opposite in direction to the force exerted on \( q \) by charge B.

### Step 3: Write the Expressions for the Forces
The force \( F_1 \) exerted on charge \( q \) by charge \( +Q \) at A is given by Coulomb's law:
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Qq}{x^2} \]
The force \( F_2 \) exerted on charge \( q \) by charge \( +Q \) at B is:
\[ F_2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Qq}{(r - x)^2} \]

### Step 4: Set the Forces Equal
For equilibrium, we set \( F_1 = F_2 \):
\[ \frac{1}{4 \pi \epsilon_0} \cdot \frac{Qq}{x^2} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Qq}{(r - x)^2} \]
Since \( \frac{1}{4 \pi \epsilon_0} \) and \( Qq \) are common factors, we can cancel them out:
\[ \frac{1}{x^2} = \frac{1}{(r - x)^2} \]

### Step 5: Cross Multiply
Cross multiplying gives us:
\[ (r - x)^2 = x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ r^2 - 2rx + x^2 = x^2 \]
Cancelling \( x^2 \) from both sides results in:
\[ r^2 - 2rx = 0 \]

### Step 7: Solve for \( x \)
Factoring out \( x \):
\[ r^2 = 2rx \]
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ r = 2x \quad \Rightarrow \quad x = \frac{r}{2} \]

### Step 8: Substitute \( x \) Back to Find \( q \)
Now, we substitute \( x = \frac{r}{2} \) back into the force equation:
\[ \frac{Qq}{\left(\frac{r}{2}\right)^2} = \frac{Q^2}{r^2} \]
This simplifies to:
\[ \frac{Qq}{\frac{r^2}{4}} = \frac{Q^2}{r^2} \]
Cross multiplying gives:
\[ 4Qq = Q^2 \]
Dividing both sides by \( Q \) (assuming \( Q \neq 0 \)):
\[ 4q = Q \quad \Rightarrow \quad q = \frac{Q}{4} \]

### Final Answer
Thus, the charge \( q \) in terms of \( Q \) is:
\[ q = \frac{Q}{4} \]