Physics answers please, appreciate it in advance.

Question

1. A positive charge of 0.00078 C is 28 m from a negative charge of 0.0006 C. What is the force of one of the charges due to the other charge in units of Newtons?
2. Charges Q1 and Q2 exert repulsive forces of 11 N on each other. What is the repulsive force in Newtons when their separation is decreased so that their final separation is 74 % of their initial separation? 
3. What is the potential energy of two charges 1.8 milliCoulombs and 4.3 milliCoulombs separated by a distance of 8 meters? Round your answer to 1 decimal place.  
4. A positive point charge of 0.00042 C is located 3 m from another positive point charge of 3.320889E-5 C. Their electric potential energy is  (Blank 1) J.
5. There is a positive point charge of 0.00036 C. Charge q1exerts a force 6.168 N on a test charge 0.062 C you bought into the electric field of q1. What is the electric field from the charge q1. Their electric field is (Blank 1) N/m.

Answer 1

Solution:

1- F = K q1q1/r² = 8.99 x10⁹ N.m²/C²[ ( 0.00078C)(0.0006C)] / (28m)² = 5.37 N

2- the charges apply repulsive force of 11 N on each other. Therefore charges are alike either both positive or both negative when the distance is 28 m. If the distance is reduced to 74% of original distance,

r = 0.74(28) = 20.72m

F = 8.99 x10⁹ N.m²/C²[ ( 0.00078C)(0.0006C)] / (20.72m)² = 9.8 N

3- The potential energy of two charges PE = k q1q2/r =

8.99 x10⁹ N.m²/C²[ ( 1.8x10^-3C)(4.3x10^-3C)] / 8m = 8698 N.m

For two positively charged charges

PE = 8.99 x10⁹ N.m²/C²[ ( 0.00042C)(3.320899x10^-5C)] / 3m = 13.9 N.m

4- Determine d² between the charge and the test charge for a force of 6.168 N

6.168N = 8.99X10⁹N.m²/C²[(0.00036C)(0.062C)] / d²

Solving for d² = 32531 m²

The value of the electric Field E = K Q / d² = 8.99X10⁹N.m²/C²(0.00036C) / 32531 m² = 99.48 N/C