Two particles A and B having charges 20µC and -5µC relatively are held fixed with a separation of 5cm. At what position a third charged particle should be placed so that it does not experience a net electric force?

Given Data:

1. Charge of particle A: q_A = +20 µC
2. Charge of particle B: q_B = -5 µC
3. Distance between A and B: r = 5 cm = 0.05 m

The third charge will experience forces due to both A and B. For the net force to be zero:

F_{A → q₀} = F_{B → q₀}

Using Coulomb's law:

k q_A q₀ / d_A² = k q_B q₀ / d_B²

Simplify (since k and q₀ cancel out):

q_A / d_A² = |q_B| / d_B²

Let d_A be the distance of the third charge from A.

Let d_B be the distance of the third charge from B.

If the third charge is placed to the right of B:

d_A = d_B + r

q_A / (d_B + r)² = |q_B| / d_B²

Substitute values:

20 / (d_B + 0.05)² = 5 / d_B²

Step 1: Simplify the equation:

4 / (d_B + 0.05)² = 1 / d_B²

Step 2: Cross-multiply:

4 d_B² = (d_B + 0.05)²

Step 3: Expand and simplify:

4 d_B² = d_B² + 0.1 d_B + 0.0025

3 d_B² - 0.1 d_B - 0.0025 = 0

Step 4: Solve using the quadratic formula:

d_B = [-b ± √(b² - 4ac)] / 2a

Where: a = 3, b = -0.1, c = -0.0025

Step 5: Substitute values and solve:

d_B = (0.1 ± √(0.01 + 0.03)) / 6

d_B = (0.1 ± 0.2) / 6

d_B = (0.1 + 0.2) / 6 = 0.05 m = 5 cm

d_B = (0.1 - 0.2) / 6 = -0.05 m (not valid)

Final Answer:

The third charged particle should be placed 5 cm to the right of -5 µC.