卷子答案网-一个不只有答案的网站4.在如图所示的四种电场中，分别标记有α，b两点。下列说法正确的是
2023~2024学年第一学期高二年级期中学业诊断
物理试卷
(考试时间：上午10：30一12：00)
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
A.甲图中的a、b两点场强和电势都相同
B.乙图中两等量异种点电荷连线的中垂线上，与连线等距的α、b两点场强和电势都相同
题号
三
四
总分
C.丙图中两等量同种点电荷连线的中垂线上，与连线等距的α、b两点电势和场强都不相同
得分
D.丁图中a、b两点场强不相同，电势相同
5.如图所示的电解池，通电1s时间内共有0.3C的正离子和0.3C的负
A
第I卷（选择题共45分）
离子通过截面MN,则电解池中MN处的电流是
A.0.5A
B.0A
一、单项选择题：本题共10小题，每小题3分，共30分。请将正确选项前字母标号填入下表内
C.0.8A
D.0.6A
相应位置。（高考不选考物理的同学请完成1~6题，每题8分，共48分）》
6.关于导体的电阻和电阻率，下列说法正确的是
题号
1
2
3
4
5
6
7
8
9
10
A.将一根导线一分为二，半根导线的电阻和电阻率都是原来的一半
答案
B降低温度时，某些金属、合金和化合物的电阻突然减小为零，这种现象叫做超导现象
C.一只白炽灯泡正常发光时灯丝的电阻为121Ω，当这只灯泡停止发光一段时间后，灯丝的
1.下列说法正确的是
电阻大于1212
A库仑定律F=k9中，k的数值是卡文迪什通过扭秤实验测得的
D.常温下，若将一根金属丝均匀拉长为原来的10倍，其电阻变为原来的10倍
7.如图所示，场源电荷-Q形成的电场中，、b是同一条电场线上的两个点。把一负试探电荷
B.元电荷e的数值，最早是由美国物理学家密立根测得的
由b点沿电场线移动到α点的过程中，下列说法正确的是
C电子的质量与电子的电荷量之比叫做电子的比荷
A.电场强度变大，电势升高
D.感应起电和摩擦起电的实质都是正电荷的转移
B.电场强度变小，电势降低
2.真空中，两个半径为R且完全相同的金属小球，分别带有+3Q和-2Q的电荷量，当它们相距
C.电场力做负功，电势能增加
为r时(>R),它们之间的库仑力大小为F。若把它们接触后分开，仍置于相距为r的两点，
D.电场力做正功，电势能减少
它们之间的库仑力大小将变为
8.如图所示，四个点电荷所带电荷量的绝对值均为Q,分别固定在正方形的四个顶点上，正方
形边长为α，静电力常量为k,则正方形两条对角线交点处的电场强度
3.关于电场，下列说法正确的是
A大小为4v2
,方向水平向左
⑧
A.电场看不见，摸不着，因此电场实际不存在
B.大小为2V2心，方向水平向左
B.处在电场中的其他电荷受到的作用力就是这个电场给予的
a2
C根据电场强度的定义式E=可知，E与F成正比，E与9成反比
C大小为4V2
,方向水平向右
a2
D.由公式E=k可知，放入电场中某点的检验电荷的电荷量Q越大，则该点的电场强度越大
D大小为2V2
r
a2
,方向水平向右
高二物理第1页（共8页）
高二物理第2页（共8页）2023～2024学年第一学期高二年级期中学业诊断
物理参考答案及评分建议
一、单项选择题：本题包含 10小题，每小题 3分，共 30分。
题号 1 2 3 4 5 6 7 8 9 10
选项 B D B B D B C A C C
二、多项选择题：本题包含 5小题，每小题 3分，共 15分。
题号 11 12 13 14 15
选项 AD BD BC CD AD
三、实验题：共 14分。
16.（6分）
（1）补画电路如图所示（2分）
（2）3.204~3.206 （2分）
3 ( )
（ ） （2分）
17.（8分）
（1）电路图如图所示（4分）
2 ( ) （ ） （4分）
四、计算题：共 41分。
18.（8分）
(1) = ·········································································· （2分）
= ···············································································（1分）
(2) = ··········································································（1分）
= ···············································································（1分）
= ··········································································（2分）
= / ·············································································（1分）
19.（8分）
（1）设绝缘线上的力为T，库仑力为F
= ················································································（1分）
= ·············································································（1分）
{#{QQABZYwEggigAhAAAQgCEwFwCgOQkAECCIoOBBAEoAAAQRFABCA=}#}
联立解得：
= ··············································································（1分）
设B球所带电荷量为q，OB线长度为l
= ····················································································（2分）
= ·················································································（1分）
联立解得：
= ··················································································（2分）
20.（12分）
（1）设第n滴在两极板间下落时作匀速直线运动
= ············································································（1分）
= ····················································································（2分）
= ·················································································（2分）
= +
联立解得 ································································· （2分）
(2)设能落在下极板的油滴数为 k
= ····················································································（1分）
+ = ········································································ （2分）
= ( + )联立解得 ·································································（2分）
21.（13分）
(1) = 经过加速电场 ····················································· （1分）
在偏转电场 = ······································································（1分）
= ··················································································（1分）
=
=
设电子出偏转电场时速度方向与 '夹角为
= ················································································ （1分）
= ··················································································· （1分）
= ···············································································（1分）
= + ··············································································（1分）
{#{QQABZYwEggigAhAAAQgCEwFwCgOQkAECCIoOBBAEoAAAQRFABCA=}#}
联立解得： = ··································································· （1分）
(2) 为使电子打到荧光屏上，设偏转电场两极板之间电压最大值为
= ··············································································· （1分）
= ·············································································（1分）
= ················································································（1分）
联立解得： = ································································（1分）
为使电子打到荧光屏上偏转电场两极板之间的
电压范围是 ≤ ≤ ······················································（1分）
{#{QQABZYwEggigAhAAAQgCEwFwCgOQkAECCIoOBBAEoAAAQRFABCA=}#}
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