Question

Two metallic balls with radiuses r = 4 cm and a bigger one with R = 8 cm are positioned at big distance from each other and all connected with a thin wire. One of the smaller balls initially was charged up to Q = 12 nCl. Two others are surrounded by metallic envelopes with radiuses r = 5 cm and R = 9 cm correspondingly. Envelopes are electrically grounded. Find the charge of the first ball after the charge equilibrium will be reached. Please, express result is nCl and write it down rounded up to 3 significant digits.

Answer 1

Look on the figure. The figure introduced more stringent non-repeating designations of the values given in the condition of the problem.

$r_1=0.04m; r_2=0.05m; R_1=0.08m; R_2=0.09m; q_1+Q+q=q_0=12nCl$ . The last equation is due to the law of conservation of charge. The total charge of the connected balls cannot change without connection to other circuit, thus the sum of charge of two small balls and a bigger one is equals to the initial charge of one small ball. At equilibrium all balls must have the null electric potential to each other, and similar electric potential U against to metallic envelopes and the ground. If this is not so the charge from one may be move to another. As the balls are positioned at big distance they don't influence on electric field of each other, and two of them with an envelopes represents a spherical condensator. The capacities of spherical condensators are

$c=4\pi\epsilon \epsilon_0\cdot(\frac{1}{r_1}-\frac{1}{r_2})^{-1}=22.3 pF$

$C=4\pi\epsilon \epsilon_0\cdot(\frac{1}{R_1}-\frac{1}{R_2})^{-1}=80.1pF$ ,

where $\epsilon$ is permittivity of dielectric, and $\epsilon _{0}=8.854×10^{−12} F⋅m^{−1}$ is the vacuum permittivity in Si unit system. The electric permittivity of air has the value $\epsilon$=1,0006, and can be neglecting as 1 when answer rounded up to 3 significant digits.

The capacitance of the 'naked' ball will be (if the ground is far enough)

$c_n=4\pi \epsilon \epsilon_0 r_1=4.45pF$

From definition of electric capacity we have $c=\frac{q_1}{U}; C=\frac{Q}{U};c_n=\frac{q}{U}$, or $U=\frac{q_1}{c}=\frac{Q}{C}=\frac{q}{c_n}$ , and $q_1=q\cdot \frac{c}{c_n} ; Q=q \cdot\frac{C}{c_n}$. When substitute this to the law of conservation charge we get

$q\cdot \frac{c}{c_n}+q \cdot\frac{C}{c_n}+ q=q_0$ , and the charge of small 'naked' ball after equilibrium will be reached is

$q=\frac{q_0}{\frac{c}{c_n}+\frac{C}{c_n}+ 1}=q_0\cdot \frac{c_n}{c+C+c_n}=12nCl\cdot\frac{4.45}{22.3+80.1+4.45}=0.5 nCl$

Answer: The charge of the first small ball after the charge equilibrium will be reached is $0.500nCl$