Two point charges q₁ = 3 μC and q₂ = –3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?

(a).

Let us imagine two points A and B such that AO=OB and O is centre of line AB.

The distance between the two charges  is AB=20cm

Therefore AO=OB=10cm

Then net electric field at point O=E (say)

Electric field at point O as a result of +3µC charge will be =E₁ (say)

Where (1/4πε₀) =9x 10⁹ and  ε₀   =Permittivity of free space

and OA =10cm =10x 10⁻²Nm ²C⁻²

Then E₁= (1/4πε₀) x 3×10⁻⁶/(OA)²

= (1/4πε₀) x 3×10⁻⁶/(10x 10⁻²)²       along OB

Electric field at point O as a result of -3µC charge will be =E2 (say)

Then magnitude of  E₂(absolute value)= (1/4πε₀) x ( -3×10⁻⁶)/(OB)²

= (1/4πε₀) x 3×10⁻⁶/(10x 10⁻²)²       along OB

Therefore E= E₁+E2= 2 x (1/4πε₀) x 3×10⁻⁶/(10x 10⁻²)²       along OB

Since E₁& E2 have electric field in the same direction it will add-up and since the magnitude of both are equal we can just double it.

Therefore E=2x 9x 10⁹ x 3×10⁻⁶/(10x 10⁻²)² NC⁻¹

=5.4 x 10⁶ NC⁻¹ along OB

Therefore, the electric field at mid-point O is 5.4 x 10⁶ NC⁻¹ along OB.

(b).

A test charge of amount 1.5 x 10^-9 C is placed at mid-point O.

q= 1.5 x 10⁻⁹C

Force experienced by the test charge = F (say)

Then F = qE

= 1.5 x 10⁻⁹ x 5.4 x 10⁶ = 8.1 x 10^-3 N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 x 10^-3 N along OA.