Two charges, Q1= 3.20 microCoulombs, and Q2=5.90 microCoulombs, are located at points (0,-2.00cm) and (0,+2.00cm.

What is the magnitude of the electric field at point P, located at (5.00cm,0), due to Q1 alone?

To find the magnitude of the electric field at point P due to charge Q1 alone, we can use the formula for the electric field intensity:

E = k * |Q1| / r^2

Where:
E is the electric field intensity,
k is Coulomb's constant (8.99 x 10^9 N m^2/C^2),
|Q1| is the magnitude of charge Q1, and
r is the distance between charge Q1 and point P.

In this case, Q1 = 3.20 microCoulombs = 3.20 x 10^-6 C,
and r = distance between Q1 and point P = 5.00 cm = 5.00 x 10^-2 m.

Substituting these values, we get:

E = (8.99 x 10^9 N m^2/C^2) * (3.20 x 10^-6 C) / (5.00 x 10^-2 m)^2

Simplifying the equation, we get:

E = (8.99 x 10^9 N m^2/C^2) * (3.20 x 10^-6 C) / (2.50 x 10^-3 m^2)

E = (8.99 x 3.20 x 10^3) / (2.50 x 10^-3) N/C

E = 11.4688 x 10^3 N/C

Therefore, the magnitude of the electric field at point P due to Q1 alone is approximately 11.5 x 10^3 N/C or 1.15 x 10^4 N/C.