In a liquid dielectric with a density of 900 kg / m3, a positively charged ball with a mass of 0.18 g is in a suspended state

In a liquid dielectric with a density of 900 kg / m3, a positively charged ball with a mass of 0.18 g is in a suspended state, the density of the ball’s substance is 1800 kg / m3, and q = 20 nC. Find the field strength in the dielectric assuming it to be uniform

To find the field strength in the considered dielectric, we project the forces onto the vertical axis: Fт = Fe + Fa; m * g = E * q + ρd * g * V = E * q + ρd * g * m / ρsh, whence we express: E = (m * g – ρd * g * m / ρsh) / q = m * g * (1 – ρд / ρш) / q.

Constants and variables: m is the mass of the ball (m = 0.18 g = 18 * 10 ^ -5 kg); g – acceleration due to gravity (g ≈ 10 m / s2); ρd – dielectric density (ρd = 900 kg / m3); ρsh – density of the ball substance (ρsh = 1800 kg / m3); q – ball charge (q = 20 nC = 2 * 10 ^ -8 C).

Let’s make a calculation: E = m * g * (1 – ρd / ρsh) / q = 18 * 10 ^ -5 * 10 * (1 – 900/1800 / (2 * 10 ^ -8) = 45 * 10 ^ 3 V / m.

Answer: The field strength in the considered dielectric is 45 kV / m.