Question

Two point charges q1 400 microcoulomb and q 200 microcoulomb at a fixed 60 cm apart in vacuum find the intensity of the electric field at midpoint of the line joining q1 and q2

Answer 1

q₁ A------------M------------B q₂
Let two charge q₁ = 400 μC and q₂ = 200 μC are placed A and B respectively and M is the midpoint of AB .
q₁ and q₂ both are positive charge . Means net electric field, E = E₁ - E₂ [ due to both the charge creates electric fleid in opposite direction as shown in figure,
Now, electric field due to q₁ = E₁ = kq₁/AM² [ AM = MB = 0.3m , from above daigram]
E₁ = 9 × 10⁹ × 400 × 10⁻⁶/(0.3)² = 4 × 10⁵ N/C

Similarly electric field due to q₂ = E₂ = Kq₂²/MB²
E₂ = 9 × 10⁹ × 200 × 10⁻⁶/(0.3)² = 2 × 10⁵ N/C

Hence, net electric field at middle point = E₁ - E₂ = 4 × 10⁵ - 2 × 10⁵ = 2 × 10⁵N/C