Answer:
The potential difference across the plates is 226 V.
Explanation:
Given;
area of the capacitor plate, A = 0.2 m²
separation, d = 0.1 mm = 0.1 x 10⁻³ m
charge on each plate, Q = 4 x 10⁻⁶ C
Charge on the capacitor is given by;
Q = CV
Where;
C is the capacitance of the capacitor, given as;
C = ε₀A / d
Then, the potential difference across the plates is given by;

Therefore, the potential difference across the plates is 226 V.
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Answer:
F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)
Explanation:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters(m)
Equivalence
1nC= 10⁻⁹C
Data
K=8.99x10⁹N*m²/C²
q₁ = 7.94-nC= 7.94*10⁻⁹C
q₂= 4.14-nC= 4.14 *10⁻⁹C
d= 1.77 m
Magnitude of the electrostatic force that one charge exerts on the other
We apply formula (1):

F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)