A ball weighing 1 kg hangs on an insulated thread, the charge of which is -5 * (10 to -9 degrees) Cl.

A ball weighing 1 kg hangs on an insulated thread, the charge of which is -5 * (10 to -9 degrees) Cl. At what distance it is necessary to bring another ball with a charge of 4.9 * (10 to -8 degrees) Cl. to double the thread tension.

m = 1 kg.

g = 10 m / s2.

q1 = -5 * 10 ^ -9 Cl.

q2 = 4.9 * 10 ^ -8 Cl.

k = 9 * 10 ^ 9 N * m2 / Cl2.

F2 = 2 * F1.

r -?

The initial tension of the thread F1 is equal to the weight of the ball: F1 = m * g.

When the second charged ball is brought up, the electrostatic force of mutual attraction F begins to act on the first ball, the value of which is determined by the Coulomb’s law: F = k * q1 * q2 / r2.

The tensile force in the second case F2 will be determined by the formula: F2 = m * g + F = m * g + k * q1 * q2 / r ^ 2.

m * g + k * q1 * q2 / r ^ 2 = 2 * m * g.

k * q1 * q2 / r ^ 2 = m * g.

r2 = k * q1 * q2 / m * g.

r = √ (k * q1 * q2 / m * g).

r = √ (9 * 10 ^ 9 N * m2 / Cl2 * 5 * 10 ^ -9 Cl * 4.9 * 10 ^ -8 Cl / 1 kg * 10 m / s2) = 4.7 * 10-4 m …

Answer: it is necessary to bring the charge closer to a distance r = 4.7 * 10 ^ -4 m.