The electric field at a point 23 cm from a charged particle is 102424 N/C toward the particle. What is the charge on the particle? (Take k=9x109 Nm2/C2 )

2.62×10⁻⁶ C

Explanation:

From the question,

Electric Field (E) = kq/r² Equation 1

Where q = charge on the particle, r = distance of the electric field from  the charge,  k = coloumb's constant.

make q the subject of the equation

q = Er²/k Equation 2

Given: E = 102424 N/C, r = 23 cm = 0.23 m, k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = (102424×0.23)/(9×10⁹)

q = 2.62×10⁻⁶ C

Hence the charge on the particle is 2.62×10⁻⁶ C

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.