(i) What is the Electric Field at the Midpoint O of the Line Ab Joining the Two Charges? Exercise Chapter 1 Electric

qA = 3 µC  = 3 × 10 ^–6  C
qB = −3 µC  =  - 3 × 10 ^–6  C
Distance AB = 20 cm = 0.2 m
In this problem, O is the mid-point of line AB.
OB = OA = 0.2/2  = 0.1 m
The electric field produced by the charge Q at a point r is given as

Plug the values we get
Electric field at point O caused by +3µC charge,

Magnitude of electric field at point O caused by −3µC charge,

The direction of electric field is always positive to negative so that
Direction of electric field will in direction of OB

Total electric field due to both charge
E _total = E₁ + E₂
 
Plug 1/4πε₀  = 9 × 10⁹ Nm²C^-2
We get
E total = 5.4 × 10⁶ N/C along OB
Answer
The electric field at mid-point O is 5.4 × 10⁶ N C⁻¹ along  the direction OB.

Given that
q = 1.5 × 10⁻⁹ C
Force experienced by the test charge = F
Use formula of electrostatic force of attraction
F = qE
Plug the values in this formula we get
F = 1.5 × 10⁻⁹ × 5.4 × 10⁶ = 8.1 × 10⁻³ N
Here placed charge is negative so negative charge at point B will repel while positive change at point A will attract. Hence the direction of force will along OA
Answer
The force experienced by the test charge = 8.1 × 10⁻³ N along the direction of OA.