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.