Answer:
(a) 1.6×10⁵ N/C
(b) -6.4×10⁻⁴ N
(c) 1.6×10⁻⁸ C
Explanation:
(a)
Electric field: Electric Field (E) is defined as the force per unit charge which it exerts at that point. Its direction is that of the force exerted on a positive charge.
mathematically,
E = F/q ......................................... Equation 1
Where E = Electric Field, F = force, q = charge.
Given: F = 8×10⁻⁴ N, q = q₀ = 5 nC = 5 ×10⁻⁹
Substituting into equation 1
E = 8×10⁻⁴/5 ×10⁻⁹
E = 1.6×10⁵ N/C.
(b)
F = Eq...................... Equation 2
Where F = force, E = electric Field, q = charge.
Given: E = 1.6×10⁵ N/C, q = -4 nC = -4×10⁻⁹ C
Substituting into equation 2
F = 1.6×10⁵× (-4×10⁻⁹)
F = -6.4×10⁻⁴ N.
(c)
q₂ = Fr²/kq₁..................... Equation 3
Where,
F = force, q₁ = first charge, q₂ = second charge, r = distance between the charges, k = proportionality constant.
Given: F = -6.4×10⁻⁴ N, q₁ = -4 nC = -4×10⁻⁹ C, r = 3 cm = 0.03 m.
Constant: k = 9×10⁹ Nm²/C²
Substituting into equation 3
q = -6.4×10⁻⁴(0.03)²/[(9×10⁹)(-4×10⁻⁹ )]
q = -5.76×10⁻⁷/(-36)
q = 1.6×10⁻⁸ C
Thus the value of charge =1.6×10⁻⁸ C
