In a horizontal uniform field with a strength of 20 kV / m, a ball with a mass of 50 g with a charge of +25 nC

In a horizontal uniform field with a strength of 20 kV / m, a ball with a mass of 50 g with a charge of +25 nC is suspended on a rubber thread with a stiffness of 6.0 N / m. The environment is vacuum. Determine the elongation of the thread on which the ball is suspended

E = 20 kV / m = 20,000 V / m.

k = 6 N / m.

m = 50 g = 0.05 kg.

g = 9.8 m / s2.

q = 25 nC = 25 * 10 ^ -9 C.

x -?

0 = m * g + N + q * E – equilibrium condition of the ball in vector form.

OH: q * E = N * sinα.

OU: m * g = N * cosα.

We express the tension force of the rubber thread by Hooke’s law: N = k * x.

q * E = k * x * sinα.

m * g = k * x * cosα.

q * E / sinα = m * g / cosα.

sinα / cosα = q * E / m * g.

tgα = q * E / m * g.

∠α = arctan (q * E / m * g).

∠α = arctan (25 * 10-9 C * 20,000 V / m / 0.05 kg * 9.8 m / s2) = 0.05 °.

k * x = m * g / cosα.

x = m * g / k * cosα.

x = 0.05 kg * 9.8 m / s2 / 6 N / m * cos0.05 ° = 0.08 m.

Answer: x = 0.08 m