When a test charge q0 = 5 nc is placed at the origin, it experiences a force of 8 10-4 n in the positive y direction.

(a) what is the electric field at the origin?

(b) what would be the force on a charge of -4 nc placed at the origin?

(c) if this force is due to a charge on the y axis at y = 3 cm, what is the value of that charge?

(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

temp rises, and expansion

in physics zero. in muscle fatigue could be high