解析广西北海市合浦一中高二下学期周练物理试卷2 含解析

解析广西北海市合浦一中高二下学期周练物理试卷2 含解析
解析广西北海市合浦一中高二下学期周练物理试卷2 含解析

2015-2016学年广西北海市合浦一中高二(下)周练物理试卷(2)

一、选择题:(每题6分,共66分)

1.关于电子伏特(eV),下列说法中正确的是()

A.电子伏特是电势的单位 B.电子伏特是电场强度的单位

C.电子伏特是能量的单位 D.1 eV=1.60×1019J

2.下列关于U=Ed说法正确的是()

A.在电场中,E跟U成正比,跟d成反比

B.对于任何电场,U=Ed都适用

C.U=Ed只适用于匀强电场,d是电场中任意两点间距离

D.U=Ed只适用于匀强电场,d是沿着电场线方向两点间距离

3.关于等势面的说法,下列哪些说法是正确的()

A.等势面和电场线处处垂直

B.同一等势面上的点场强大小必定处处相等

C.电荷所受电场力的方向必和该点等势面垂直,并指向电势升高的方向

D.在同一等势面上移动电荷,电场力一定不做功

4.关于电场力做功和电势差的说法中,正确的是()

A.电势差的大小由电场力在两点间移动电荷做的功和电荷量决定

B.电场力在电场中两点间移动电荷做功的多少由这两点间的电势差和电荷量决定

C.电势差是矢量,电场力做的功是标量

D.在匀强电场中,与电场线垂直的某个方向上任意两点间的电势差均为零

5.下列措施中,不属于防止静电危害的是()

A.油罐车后面有一条拖在地面上的铁链

B.小汽车上有一根露在外面的小天线

C.在印染厂中保持适当的湿度

D.在地毯中夹杂0.05mm~0.07mm的不锈钢丝导电纤维

6.某电场的电场线分布如图所示,以下说法正确的是()

A.c点场强大于b点场强

B.a点电势小于b点电势

C.若将一试探电荷+q由a点释放,它将沿电场线运动到b点

D.若在d点再固定一点电荷﹣Q,然后将一试探电荷+q由a移至b的过程中,电势能减小7.在电场中A、B两点间的电势差为U AB=75V,B、C两点间的电势差为U BC=﹣200V,

则A、B、C三点电势高低关系为()

A.φA>φB>φC B.φA<φC<φB C.φC>φA>φB D.φC>φB>φA

8.在一正点电荷电场中A、B两点间的电势差U AB=200V,将电量为+2×10﹣8C的电荷q

从A点移到B点,则下列说法中错误的是()

A.电场力对电荷做正功4×10﹣6J

B.电荷q具有的电势能增大4×10﹣6J

C.电荷q的动能增大了4×10﹣6J

D.电荷q电势能的减小量等于动能的增加量

9.如图所示,是一个正电荷的电场,两虚线是以点电荷为圆心的同心圆,A、B两点的电势差为4V.有一个q=﹣1.0×10﹣8C的点电荷,从A点沿不规则曲线移到B点,电场力做功为()

A.4.0×10﹣8J B.﹣4.0×10﹣8J

C.﹣2.5×10﹣9J D.路径不清,不能确定

10.如图所示,a、b、c是一条电场线上的三个点,电场线的方向由a到c.a、b间的距离等于b、c间的距离.用φa、φb、φc和E a、E b、E c分别表示三点的电势和场强,可以判定()

A.φa﹣φb=φb﹣φc B.φa>φb>φc C.E a>E b>E c D.E a=E b=E c

11.如图所示表示某静电场等势面的分布,电荷量为1.6×10﹣9C的正电荷从A经B、C到达D点.则从A到D电场力对电荷做的功为()

A.4.8×10﹣8J B.﹣4.8×10﹣8J C.8.0×10﹣8J D.﹣8.0×10﹣8J

二、计算题:(共34分)

12.如图所示的匀强电场中,有a、b、c三点,ab=5cm,bc=12cm,其中ab沿电场方向,bc和电场方向成60°角,一个电荷量为q=4×10﹣8C的正电荷从a移到b电场力做功为W1=1.2×10﹣7J求:

(1)匀强电场的场强E

(2)电荷从b移到c,电场力做功W2

(3)a、c两点的电势差U ac.

2015-2016学年广西北海市合浦一中高二(下)周练物理

试卷(2)

参考答案与试题解析

一、选择题:(每题6分,共66分)

1.关于电子伏特(eV),下列说法中正确的是()

A.电子伏特是电势的单位 B.电子伏特是电场强度的单位

C.电子伏特是能量的单位 D.1 eV=1.60×1019J

【考点】元电荷、点电荷.

【分析】本题考查学生对单位的理解,知道电子伏是能量的单位.

【解答】解:A、电子伏特是能量的单位,代表一个电子电位改变(增加或减少)单位伏特时其能量的改变量;故AB错误,C正确;

D、1eV=1.60×10﹣19J,故D错误;

故选:C.

2.下列关于U=Ed说法正确的是()

A.在电场中,E跟U成正比,跟d成反比

B.对于任何电场,U=Ed都适用

C.U=Ed只适用于匀强电场,d是电场中任意两点间距离

D.U=Ed只适用于匀强电场,d是沿着电场线方向两点间距离

【考点】匀强电场中电势差和电场强度的关系.

【分析】(1)在电场中,电场强度E是由电场本身决定的,与其他的因素无关;

(2)U=Ed是根据在匀强电场中移动点电荷时,电场力做功与电势差的关系导出的公式,因此只适用于匀强电场,d是沿着电场线方向两点间距离,若不是沿着电场线方向,则使用公式U=Edcosθ.

【解答】解:A:在电场中,电场强度E是由电场本身决定的,与其他的因素无关,故A 错误;

BCD:U=Ed是根据在匀强电场中移动点电荷时,电场力做功与电势差的关系导出的公式,因此只适用于匀强电场,d是沿着电场线方向两点间距离,若不是沿着电场线方向,则使用公式U=Edcosθ,故BC错误,D正确.

故选:D

3.关于等势面的说法,下列哪些说法是正确的()

A.等势面和电场线处处垂直

B.同一等势面上的点场强大小必定处处相等

C.电荷所受电场力的方向必和该点等势面垂直,并指向电势升高的方向

D.在同一等势面上移动电荷,电场力一定不做功

【考点】等势面.

【分析】电场中电势相等的各个点构成的面叫做等势面;等势面与电场线垂直,沿着等势面移动点电荷,电场力不做功.场强与电场线的分布情况有关.正电荷所受电场力与场强方向相同,负电荷所受电场力与场强方向相反.由此分析即可.

【解答】解:A、等势面与电场线的关系是:等势面和电场线处处垂直,故A正确.

B、电场中电势相等的各个点构成的面叫做等势面,等势面上各个点的场强大小情况要看具体的电场,场强不一定处处相等,故B错误;

C、负电荷所受电场力的方向必和该点等势面垂直,并指向电势升高的方向,而正电荷所受电场力的方向和该点等势面垂直,并指向电势降低的方向.故C错误.

D、电荷所受的电场力方向与等势面垂直,则在同一等势面上移动电荷时,电场力与速度方向一直垂直,电场力不做功,故D正确;

故选:AD

4.关于电场力做功和电势差的说法中,正确的是()

A.电势差的大小由电场力在两点间移动电荷做的功和电荷量决定

B.电场力在电场中两点间移动电荷做功的多少由这两点间的电势差和电荷量决定

C.电势差是矢量,电场力做的功是标量

D.在匀强电场中,与电场线垂直的某个方向上任意两点间的电势差均为零

【考点】匀强电场中电势差和电场强度的关系.

【分析】电势差的定义公式:U AB=,是用比值法定义的,电势差与试探电荷无关,电

势差是标量;在匀强电场中,与电场线垂直的某个方向上为等势面.

【解答】解:A、电场中两点间的电势差与试探电荷无关,由电场本身和两点的位置共同决定,故A错误;

B、根据公式W AB=qU AB,电场力在两点间移动电荷做功的多少由两点间的电势差和该电荷的电荷量决定,故B正确;

C、公式:U AB=,是用比值法定义的,可知电势差是标量,电场力做的功是标量.故C

错误;

D、由于等势面与电场线垂直,所以在匀强电场中,与电场线垂直的某个方向上任意两点间的电势差均为零,故D正确;

故选:BD.

5.下列措施中,不属于防止静电危害的是()

A.油罐车后面有一条拖在地面上的铁链

B.小汽车上有一根露在外面的小天线

C.在印染厂中保持适当的湿度

D.在地毯中夹杂0.05mm~0.07mm的不锈钢丝导电纤维

【考点】静电现象的解释.

【分析】静电危害是由于相互间不断摩擦,从而产生大量的静电,不及时导走,会出现放电危害.

【解答】解:A、油罐车在运输过程中,油和罐之间不断摩擦,从而产生大量的静电,通过后面装一条拖地的铁链,及时导走,这是防止静电危害.不符合题意.

B、小汽车上有一根露在外面的小天线,无线信号接收天线.符合题意.

C、印刷车间中,纸张间摩擦产生大量静电,所以印刷车间中保持适当的湿度,及时把静电导走,避免静电造成的危害.不符合题意.

D、不锈钢丝的作用是把鞋底与地毯摩擦产生的电荷传到大地上,以免发生静电危害.不符合题意.

本题选择不属于防止,故选:B.

6.某电场的电场线分布如图所示,以下说法正确的是()

A.c点场强大于b点场强

B.a点电势小于b点电势

C.若将一试探电荷+q由a点释放,它将沿电场线运动到b点

D.若在d点再固定一点电荷﹣Q,然后将一试探电荷+q由a移至b的过程中,电势能减小【考点】电场线;电场强度;电势能.

【分析】电场线的疏密表示电场的强弱,沿电场线方向电势逐渐降低.判断电势能的高低可通过电场力做功来判断.注意电场线不一定与带电粒子的运动轨迹重合,除非电场线是直线,且初速度为0或与电场线平行.

【解答】解:A.电场线的疏密表示电场的强弱,b点比c点密,所以c点场强小于b点场强.故A错误.

B.沿电场线方向电势逐渐降低,所以a点电势大于b点电势.故B错误.

C.由于电场线是曲线,将一试探电荷+q由a点释放,运动轨迹不可能与电场线重合.故C错误.

D.在d点再固定一点电荷﹣Q,然后将一试探电荷+q由a移至b的过程中,电场力做正功,电势能减少.故D正确.

故选D.

7.在电场中A、B两点间的电势差为U AB=75V,B、C两点间的电势差为U BC=﹣200V,则A、B、C三点电势高低关系为()

A.φA>φB>φC B.φA<φC<φB C.φC>φA>φB D.φC>φB>φA

【考点】电势差.

【分析】本题根据电势差与电势的关系:U AB=φA﹣φB,U BC=φB﹣φC,U AC=U AB+U BC,即可判断三点电势的高低关系.

【解答】解:由题意,U AB=φA﹣φB=75V,则得:φA>φB;

U BC=φB﹣φC=﹣200V,则得:φB<φC;

又U AC=U AB+U BC=(75﹣200)V=﹣125V,则得:φA<φC;

故有:φC>φA>φB;故ABD错误,C正确.

故选:C.

8.在一正点电荷电场中A、B两点间的电势差U AB=200V,将电量为+2×10﹣8C的电荷q

从A点移到B点,则下列说法中错误的是()

A.电场力对电荷做正功4×10﹣6J

B.电荷q具有的电势能增大4×10﹣6J

C.电荷q的动能增大了4×10﹣6J

D.电荷q电势能的减小量等于动能的增加量

【考点】电势能.

【分析】根据公式W AB=qU AB求解电场力做功;电场力做正功时,电荷的电势能将减小,相反,电势能将增加;根据动能定理分析动能的变化量.

【解答】解:

A、电场力对电荷做功为W AB=qU AB=2×10﹣8×200J=4×10﹣6J,是正功,故A正确.

B、电场力对电荷做正功4×10﹣6J,则其电荷电势能减小4×10﹣6J,故B错误.

C、根据动能定理:合外力对物体做的总功等于物体动能的变化,由于不知是否有其他力做功,所以不能确定动能的增加量,则电荷q的动能不一定增大4×10﹣6J,故C错误.

D、由上分析可知,电荷q电势能的减小量不一定等于动能的增加量,故D错误.

本题选错误的,故选:BCD

9.如图所示,是一个正电荷的电场,两虚线是以点电荷为圆心的同心圆,A、B两点的电势差为4V.有一个q=﹣1.0×10﹣8C的点电荷,从A点沿不规则曲线移到B点,电场力做功为()

A.4.0×10﹣8J B.﹣4.0×10﹣8J

C.﹣2.5×10﹣9J D.路径不清,不能确定

【考点】电势差与电场强度的关系.

【分析】电场力做功与重力做功的特点类似,只与始末位置有关,与路径无关,根据公式W=qU列式求解即可.

【解答】解:由电场力做功的公式:W=qU=﹣1.0×10﹣8C×4=﹣4.0×10﹣8J

故选:B.

10.如图所示,a、b、c是一条电场线上的三个点,电场线的方向由a到c.a、b间的距离等于b、c间的距离.用φa、φb、φc和E a、E b、E c分别表示三点的电势和场强,可以判定()

A.φa﹣φb=φb﹣φc B.φa>φb>φc C.E a>E b>E c D.E a=E b=E c

【考点】电势.

【分析】本题根据顺着电场线方向电势逐渐降低,判断电势关系;电场线的疏密表示电场强度的相对大小.根据匀强电场中场强与电势差的关系U=Ed,定性分析电势差的关系.

【解答】解:

ACD、只有匀强电场中U ab=φa﹣φb=Ed ab,这个关系成立,其余电场由于E是变化的,故这个关系不成立,该题只有一条电场线,不能确定电场线的分布情况,无法比较场强的大小,故无法比较两点间电势差关系,故A错误,C错误,D错误.

B、沿电场线方向电势降低,可以比较电势高低,根据电场线方向可知φa>φb>φc,故B

正确.

故选:B.

11.如图所示表示某静电场等势面的分布,电荷量为1.6×10﹣9C的正电荷从A经B、C到达D点.则从A到D电场力对电荷做的功为()

A.4.8×10﹣8J B.﹣4.8×10﹣8J C.8.0×10﹣8J D.﹣8.0×10﹣8J

【考点】等势面.

【分析】根据W=qU可知,电场力做功和具体路径无关,只与初末位置的电势差有关,由此可正确解答.

【解答】解:A与D之间的电势差为:U AD=﹣40V﹣(﹣10V)=﹣30V.

W=qUAD=﹣4.8×10﹣8J,故ACD错误,B正确.

故选:B.

二、计算题:(共34分)

12.如图所示的匀强电场中,有a、b、c三点,ab=5cm,bc=12cm,其中ab沿电场方向,bc和电场方向成60°角,一个电荷量为q=4×10﹣8C的正电荷从a移到b电场力做功为W1=1.2×10﹣7J求:

(1)匀强电场的场强E

(2)电荷从b移到c,电场力做功W2

(3)a、c两点的电势差U ac.

【考点】匀强电场中电势差和电场强度的关系;电势能.

【分析】(1)根据电场力做功公式W=qEd,求解电场强度,d是电场线方向两点间的距离.(2)电场力做功公式W=qEd,求解电荷从b移到c电场力做功W2.

(3)先求出电荷从a到c电场力做功,再求解a、c两点的电势差U ac.

【解答】解:(1)由题,由W1=qEl ab得

E==

(2)电荷从b移到c电场力做功为

W2=qEl bc cos60°=4×10﹣8×60×0.12×0.5J=1.44×10﹣7J

(3)电荷从a移到c电场力做功为

W ac=W1+W2

则a、c两点的电势差为Uac===6.6V.

答:(1)匀强电场的场强E=60V/m.

(2)电荷从b移到c电场力做功W2=1.44×10﹣7J.

(3)a、c两点的电势差U ac=6.6V.

2016年10月7日

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