Given:
- Charge 1 (q1) = 4 microcoulomb (4 x 10^-6 C)
- Charge 2 (q2) = -16 microcoulomb (-16 x 10^-6 C)
- Distance between the charges (d) = 1 meter
- Permittivity of air (ε0) = 8.854 x 10^-12 C^2/Nm^2

Formula:
The electric intensity (E) at a point due to a point charge is given by Coulomb's law:
E = (k * q) / r^2

Where:
- k = 1 / (4πε0) (Coulomb's constant)
- q = charge
- r = distance from the charge

Calculating the Electric Intensity:

Step 1: Calculate the electric field intensity due to the first charge (q1) at the midpoint.
Using Coulomb's law:
E1 = (k * q1) / r1^2
= (k * 4 x 10^-6) / (0.5)^2 [As the midpoint is half the distance between the charges]
= (9 x 10^9 * 4 x 10^-6) / (0.5)^2
= 72 x 10^3 N/C

Step 2: Calculate the electric field intensity due to the second charge (q2) at the midpoint.
Using Coulomb's law:
E2 = (k * q2) / r2^2
= (k * -16 x 10^-6) / (0.5)^2 [As the midpoint is half the distance between the charges]
= (9 x 10^9 * -16 x 10^-6) / (0.5)^2
= -288 x 10^3 N/C

Step 3: Calculate the resultant electric field intensity at the midpoint.
The electric field intensity due to each charge is a vector quantity. Since the charges are of opposite sign, their electric field intensities will have opposite directions.
The resultant electric field intensity (EResultant) at the midpoint is the vector sum of E1 and E2.
EResultant = E1 + E2
= 72 x 10^3 N/C - 288 x 10^3 N/C
= -216 x 10^3 N/C

Conclusion:
The resultant electric intensity at the midpoint between the two charges is -216 x 10^3 N/C. The negative sign indicates that the electric field intensity is directed towards the charge with a magnitude of 216 x 10^3 N/C.

Tq I got it