Answer:
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
