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(一)两种解法受1986年高考附加题的启示,笔者在课外布置学生一题:一个厚度为d,面积为S的空气平行板电容器,其电容为C_0,现插入一块厚度为d/2、面积为S/2,介电常为ε的电介质(如图1),由此所组成的电容器的电容为多大. 学生解题的方法有如下两种. 第一种解法:如图2所示把电容器一拆为三,C_1为面积为S/2、厚度为d/2的空气电容器;C_2为面积为S/2、厚度为d/2的充满电介质的电容器;C_3为面积为S/2、厚度为d的空气电容器,其电容大小分别为C_1=C_0,C_2=εC_0,C_3=1/2C_0·求算组合电容器的电容时,先把C_1与C_2串联,再与C_3并联.
(a) The two solutions were inspired by the 1986 college entrance exam questions. The author placed a student outside the classroom: an air-parallel plate capacitor with a thickness of d and an area of S. Its capacitance is C_0, and a thickness of d/2 is inserted. The area is S/2, dielectric is often ε dielectric (Figure 1), the capacitance of the capacitor formed by this is how much. Students solve the problem in the following two methods. The first solution: as shown in Figure 2 Shows that the capacitor is split into three, C_1 is an air capacitor with an area of S/2 and a thickness of d/2; C_2 is a capacitor filled with a dielectric with an area of S/2 and a thickness of d/2; C_3 is an area of S/. 2. For an air capacitor of thickness d, the capacitances are C_1=C_0, C_2=εC_0, and C_3=1/2C_0. To calculate the capacitance of the combined capacitor, first connect C_1 in series with C_2, and then connect in parallel with C_3.