Question

Два металлических шарика, имеющих заряды Q1 = 0,2 нКл и Q2 = 0,4 нКл, привели в соприкосновение и разъединили. Найдите силу взаимодействия между ними на расстоянии 10 см. Среда - воздух.

Answer 1

Ответ:

заряд шаров после соприкосновения Q1=Q2=0.3 Кл

F=(k*Q1*Q2)/r^2=(9*10^9*0.09*10^-18)/0.01=81*10^-9

Answer 2

Answer in detail. Two metal balls, having charges Q1 = 0.2 nC and Q2 = 0.4 nC, were brought into contact and separated. Find the force of interaction between them at a distance of 10 cm. The medium is air.

Here is a detailed solution:

- When two metal balls with charges Q1 and Q2 are brought into contact, they will share their charges equally, since they are conductors. Therefore, the final charges on each ball will be Q = (Q1 + Q2) / 2 = (0.2 + 0.4) / 2 nC = 0.3 nC. - The force of interaction between two point charges Q1 and Q2 separated by a distance r is given by Coulomb's law: F = k * Q1 * Q2 / r^2, where k is the Coulomb's constant, which depends on the permittivity of the medium. In air, k = 9 * 10^9 N * m^2 / C^2. - Substituting the values of Q1, Q2, k and r in the formula, we get: F = 9 * 10^9 * 0.3 * 10^-9 * 0.3 * 10^-9 / (0.1)^2 N = 8.1 * 10^-9 N. - The force is attractive, since the charges have the same sign, and it is directed along the line joining the centers of the balls.

Therefore, the force of interaction between the two metal balls at a distance of 10 cm is 8.1 * 10^-9 N, and it is attractive. You can find more information about Coulomb's law and electric charges [here](http://du-july.ru/node/37).