Threee charges `+Q_(1),+Q_(2)` and q are placed on a straight line such that q is somewhere in between `+Q_(1)`and `Q_(2)`. If this system of charges is in equlibirium what should be the magnitude and sign of charge q?

Options

A. `(Q_(1)Q_(2))/((sqrt(Q_(1))+sqrt(Q_(2)))^(2))`, positive
B. `(Q_(1)+Q_(2))/(2)`, positive
C. `(Q_(1)Q_(2))/((sqrt(Q_(1))+sqrtQ_(2))^(2))`, Negative
D. `(Q_(1)+Q_(2))/(2)`,Negative

Explanation

To solve the problem of finding the magnitude and sign of charge \( q \) placed between two positive charges \( +Q_1 \) and \( +Q_2 \) such that the system is in equilibrium, we can follow these steps:

### Step 1: Understand the Forces Acting on Charge \( q \)
Since \( +Q_1 \) and \( +Q_2 \) are both positive charges, they will repel each other. The charge \( q \) must be negative for it to be attracted to both \( +Q_1 \) and \( +Q_2 \) to maintain equilibrium. Therefore, we conclude that:
- **Charge \( q \) must be negative.**

### Step 2: Set Up the Equilibrium Condition
For the charge \( q \) to be in equilibrium, the net force acting on it must be zero. The forces acting on \( q \) due to \( +Q_1 \) and \( +Q_2 \) can be expressed as:
- The force exerted by \( +Q_1 \) on \( q \) (attractive force) is given by:
\[ F_1 = k \frac{Q_1 |q|}{x^2} \]
- The force exerted by \( +Q_2 \) on \( q \) (also attractive) is given by:
\[ F_2 = k \frac{Q_2 |q|}{(L - x)^2} \]

Where:
- \( k \) is Coulomb's constant,
- \( x \) is the distance from \( +Q_1 \) to \( q \),
- \( L \) is the total distance between \( +Q_1 \) and \( +Q_2 \).

### Step 3: Equate the Forces
For equilibrium, we set the magnitudes of these forces equal to each other:
\[ F_1 = F_2 \]
This gives us:
\[ k \frac{Q_1 |q|}{x^2} = k \frac{Q_2 |q|}{(L - x)^2} \]
We can cancel \( k \) and \( |q| \) (since \( q \) is negative, we consider the magnitude):
\[ \frac{Q_1}{x^2} = \frac{Q_2}{(L - x)^2} \]

### Step 4: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ Q_1 (L - x)^2 = Q_2 x^2 \]
Expanding this leads to:
\[ Q_1 (L^2 - 2Lx + x^2) = Q_2 x^2 \]
Rearranging gives:
\[ Q_1 L^2 - 2Q_1 Lx + (Q_1 - Q_2)x^2 = 0 \]

### Step 5: Solve the Quadratic Equation
This is a quadratic equation in \( x \):
\[ (Q_1 - Q_2)x^2 - 2Q_1 Lx + Q_1 L^2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = Q_1 - Q_2 \),
- \( b = -2Q_1 L \),
- \( c = Q_1 L^2 \).

### Step 6: Find the Magnitude of Charge \( q \)
From the equilibrium condition, we can express the magnitude of charge \( q \):
\[ |q| = \frac{Q_1 Q_2}{\left(\frac{Q_1}{Q_2} \cdot (L - x)^2\right)} \]
Substituting \( x \) from the previous step will give us the magnitude of \( q \).

### Final Result
Thus, the magnitude and sign of charge \( q \) is:
\[ q = -\frac{Q_1 Q_2}{\left(\sqrt{Q_1} + \sqrt{Q_2}\right)^2} \]