Let the capacitance between A and B (which can be regarded as a parallel plate capacitor) be C 1, and the capacitance between C and B be C2. According to the capacitance formula of parallel plate capacitor, the capacitance is inversely proportional to the distance between plates, as follows:

c 1 = 2 * C2 & lt； 1 & gt；

The potential difference between B and A is equal to that between B and C (because the potentials of A and C are equal, B itself is an equipotential body), and it is set to u:

For the capacitance between a and c: U=Q 1/C 1, for the capacitance between b and c: U=Q2/C2, so

Q 1/C 1=Q2/C22 & gt where q1and Q2 are the charges inside A and C, respectively, depending on the conditions.

Q 1+Q2 = Q & lt； 3 & gtQ=2uC

Synthesizing the above formulas, we can get the following results: Q 1=2uC, Q2= 1uC.

So the answer is that the charged side near A of B is equal to the absolute value of the charged inside A (opposite sign), both of which are 2uC；;

The absolute charged values of the other side of B and the inner side of C are1UC; ; The outer sides of a and c are not charged.