In the circuit shown in the figure, in order to make the upper plate of capacitor C still positively charged and increase the amount of charge, the following methods can be adopted.

When balancing, the voltage across the capacitor has nothing to do with the capacitor itself, but is determined by four resistors.

Let R 1 and R2 be points A, and R3 and R4 be points B. ..

Since R 1 and R2 are connected in series, R3 and R4 are connected in parallel, the potentials at point A and point B are relatively independent (assuming the negative potential of the battery is 0):

The potential at point A is determined by R 1 and R2-φ a = u (r1) = e * r1/(r1+R2).

The potential at point B is determined by R3 and R4 -φ b = u (R3) = e * R3/(R3+R4).

The voltage on the capacitor u (c) = φ a-φ b.

If the plate is still positively charged and the amount of charge increases (Q=CU), it is necessary to increase the voltage U(C) across the capacitor, which can be achieved by increasing φA or decreasing φ B. ..

Analyze this item by item:

①. φA = e * R 1/(r1+R2) = e * [1-R2/(R 1+R2)]. When r1increases, R2/(r

②.φA=E*R 1/(R 1+R2)。 After increasing R2, R 1/(R 1+R2) becomes smaller, and φA becomes smaller.

③. φB = e * R3/(R3+R4) = e * [1-R4/(R3+R4)]. When R3 increases, R4/(R3+R4) decreases and φ b increases.

④ φB = e * R3/(R3+R4). When R4 increases, R3/(R3+R4) decreases and φ b decreases.

So only ① ④ conforms to our previous analysis, and C is selected.