A certain charge Q is divided at first into two parts, q and Q-q. later on the charges are placed at a certain distance. If he force of interaction between two charges is maximum, then

Options

A. `Q/q=4/1`
B. `Q/q=2/1`
C. `Q/q=3/1`
D. `Q/q=1/3`

Answer

To solve the problem, we need to find the ratio \( \frac{Q}{q} \) such that the force of interaction between the two charges \( q \) and \( Q - q \) is maximized. Here’s a step-by-step solution:

### Step 1: Understand the Force Between Charges
The force \( F \) between two point charges \( q \) and \( Q - q \) separated by a distance \( d \) is given by Coulomb's law:
\[ F = k \frac{q (Q - q)}{d^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Express the Force in Terms of \( q \)
Substituting \( Q - q \) into the force equation, we have:
\[ F = k \frac{q (Q - q)}{d^2} = k \frac{(qQ - q^2)}{d^2} \]

### Step 3: Differentiate the Force with Respect to \( q \)
To find the maximum force, we need to differentiate \( F \) with respect to \( q \) and set the derivative equal to zero:
\[ \frac{dF}{dq} = k \frac{(Q - 2q)}{d^2} \]
Setting the derivative to zero for maximization:
\[ Q - 2q = 0 \]
From this, we can solve for \( q \):
\[ Q = 2q \]

### Step 4: Find the Ratio \( \frac{Q}{q} \)
Now, we can find the ratio \( \frac{Q}{q} \):
\[ \frac{Q}{q} = \frac{2q}{q} = 2 \]

### Step 5: Verify Maximum Condition
To ensure that this value gives a maximum, we can check the second derivative:
\[ \frac{d^2F}{dq^2} = k \frac{-2}{d^2} \]
Since \( k \) and \( d^2 \) are both positive, \( \frac{d^2F}{dq^2} < 0 \), confirming that the force is indeed maximized when \( \frac{Q}{q} = 2 \).

### Final Answer
Thus, the ratio \( \frac{Q}{q} \) that maximizes the force of interaction between the two charges is:
\[ \frac{Q}{q} = 2 \]