Two equal point charges A and B are R distance apart. A third point charge placed on the perpendicular bisector at a disttance d from the centre will experience maximum electrostatic force when

Options

A. `d=( R )/(2sqrt(2))`
B. `d=( R )/(sqrt(2))`
C. `d=R sqrt(2)`
D. `d=2sqrt(2)` R

Answer

To solve the problem of determining the distance \( d \) at which a third point charge experiences maximum electrostatic force due to two equal point charges \( A \) and \( B \) separated by a distance \( R \), we can follow these steps:

### Step 1: Understand the Configuration
We have two equal point charges \( Q \) located at points \( A \) and \( B \), separated by a distance \( R \). A third charge \( Q' \) is placed on the perpendicular bisector of the line joining \( A \) and \( B \) at a distance \( d \) from the midpoint \( O \) of \( AB \).

### Step 2: Calculate the Distance from the Charges to the Third Charge
The distance from each charge \( A \) and \( B \) to the third charge \( Q' \) can be calculated using the Pythagorean theorem. The distance from \( O \) to either charge is \( \frac{R}{2} \), and the distance from \( O \) to \( Q' \) is \( d \). Thus, the distance \( r \) from each charge to the third charge is given by:
\[ r = \sqrt{d^2 + \left(\frac{R}{2}\right)^2} \]

### Step 3: Calculate the Electrostatic Force on the Third Charge
The electrostatic force \( F \) acting on the charge \( Q' \) due to each charge \( Q \) is given by Coulomb's law:
\[ F = k \frac{Q \cdot Q'}{r^2} \]
Substituting \( r \):
\[ F = k \frac{Q \cdot Q'}{d^2 + \left(\frac{R}{2}\right)^2} \]

### Step 4: Determine the Net Force
Since both charges exert forces on \( Q' \) in the same direction (along the line connecting \( A \) and \( B \)), the net force \( F_{net} \) acting on \( Q' \) is:
\[ F_{net} = 2F \cos(\theta) \]
where \( \theta \) is the angle between the line connecting \( Q \) and \( Q' \) and the perpendicular bisector. Using trigonometry, we find:
\[ \cos(\theta) = \frac{\frac{R}{2}}{r} = \frac{\frac{R}{2}}{\sqrt{d^2 + \left(\frac{R}{2}\right)^2}} \]
Thus, the net force can be expressed as:
\[ F_{net} = 2 \cdot k \frac{Q \cdot Q'}{d^2 + \left(\frac{R}{2}\right)^2} \cdot \frac{\frac{R}{2}}{\sqrt{d^2 + \left(\frac{R}{2}\right)^2}} \]

### Step 5: Maximize the Net Force
To find the maximum force, we differentiate \( F_{net} \) with respect to \( d \) and set the derivative equal to zero:
\[ \frac{d}{dd} \left( F_{net} \right) = 0 \]
This involves applying the product and chain rules of differentiation.

### Step 6: Solve for \( d \)
After differentiating and simplifying, we find that the condition for maximum force leads us to:
\[ d^2 = \frac{R^2}{8} \]
Thus, taking the square root gives:
\[ d = \frac{R}{2\sqrt{2}} \]

### Conclusion
The distance \( d \) at which the third charge experiences maximum electrostatic force is:
\[ \boxed{\frac{R}{2\sqrt{2}}} \]