Medan listrik dan magnetik
International Baccalaureate · IB Diploma · Physics · SL Topik 14 4:18 Narasi bahasa Inggris · Subtitle bahasa Inggris + 中文 disematkan langsung
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A negatively charged particle experiences an electric force opposite to the field direction.
带负电的粒子受到与电场方向相反的电场力。
Field direction is defined using a positive test charge 试探电荷.
电场方向是用正试探电荷来定义的。
Electric field strength is force per unit positive test charge.
电场强度是单位正试探电荷所受的力。
Like charges repel and unlike charges attract.
同种电荷相互排斥，异种电荷相互吸引。
Field lines point away from positive charges and toward negative charges.
电场线从正电荷出发指向负电荷。
Closely spaced lines represent a stronger field within a consistently drawn diagram.
在绘制一致的图中，线条密集处表示场强较强。
Electric field strength is electric force per unit positive test charge.
电场强度：单位正试探电荷所受的电场力。
A test charge is a small charge used to probe an electric field.
试探电荷：用于探测电场的微小电荷。
In a uniform field between ideal parallel plates, E equals V divided by d, away from edge effects.
在理想平行板之间的均匀电场中，E等于V除以d，忽略边缘效应。
The force is F equals q times E, so its direction reverses for a negative charge.
作用力F等于q乘以E，因此负电荷受力方向相反。
Coulomb force between ideal point charges decreases with separation squared.
理想点电荷间的库仑力随距离平方成反比减小。
In the lab: Sketch labelled plate polarities, field arrows and the charge before calculating.
在实验室中，计算前先画出标注极性的极板、电场箭头和电荷。
Convert separation to metres.
将间距转换为米。
In school, use simulations or approved low-voltage electrostatic models rather than exposed high-voltage equipment.
在学校里，请使用模拟软件或经批准的低压静电模型，而不是裸露的高压设备。
Known: plate potential difference is 120 V and separation is 0.020 m.
已知：极板间电势差为 120 V，间距为 0.020 m。
E equals V divided by d, which equals 6,000 V per m.
电场强度 E 等于 V 除以 d，即 6,000 V/m。
A plus 2.0 microcoulomb charge experiences F equals q times E, which equals 0.012 N along the field; a minus 2.0 microcoulomb charge experiences the same magnitude opposite to it.
带 +2.0 μC 电荷的粒子受到的力 F 等于 q 乘 E，即 0.012 N，方向沿电场方向；带 −2.0 μC 电荷的粒子受到的力大小相同，方向相反。
Your turn: Use E = V/d.
轮到你了。
Find E for 100 V across 0.025 m.
使用公式 E = V / d。 求 100 V 电压、间距为 0.025 m 时的电场强度 E。
Pause the video and work it out.
请暂停视频，自己算一算。
Check: 100 volts divided by 0.025 metres is 4,000 volts per metre.
核对：100 伏 ÷ 0.025 米，得每米 4000 伏。
Field strength is a vector.
场强是矢量。
Potential difference is energy per charge and is not the same quantity.
电势差是单位电荷的能量，与场强不是同一物理量。
A field-line drawing is a model, not a set of physical threads.
电场线图是一种模型，并非真实的物理线。
Not true: Every charged particle experiences a force along the electric field direction.
不正确。 并非所有带电粒子受到的力都沿电场方向。
Field direction is defined by the force on a POSITIVE charge.
电场方向定义为正电荷所受的力的方向。
A negative particle feels a force opposite to the field, so the field direction is not every particle's force direction.
负电荷受到的力与电场方向相反，因此电场方向并不等同于所有粒子的受力方向。
Electric field strength is force per unit positive test charge.
电场强度是单位正试探电荷所受的力。
In a uniform field between ideal parallel plates, E equals V divided by d, away from edge effects.
在理想平行板间的匀强电场中，E等于V除以d，忽略边缘效应。
Watch for: field strength is a vector.
注意：电场强度是矢量。
A potential difference of 60 volts is across a gap of 0.020 metres.
0.020米间距两端的电势差为60伏。
Find the electric field strength.
求电场强度。
A negative charge feels the opposite force.
负电荷受到的力方向相反。
The result is a field of 3000 volts per metre.
结果是电场强度为每米3000伏。
Field lines point from positive to negative.
电场线从正电荷指向负电荷。
Between parallel plates they are straight and evenly spaced, so the force qE is the same everywhere, and a negative charge is pushed the opposite way.
平行板之间电场线笔直且间距均匀，所以各处的力qE相同，负电荷受力方向相反。
Parallel plates have 120 V across 0.030 m.
平行板之间电压为120 V，间距为0.030 m。
Find field and force on a −2.0 microcoulomb charge.
求电场强度及作用在负2.0微库仑电荷上的力。
State directions and the uniform-field boundary.
说明方向及匀强电场的边界条件。
E is 4,000 volts per metre, from the positive to the negative plate.
电场强度 E 等于 4000 V/m，方向从正极板指向负极板。
The force magnitude is 0.0080 N, opposite the field.
作用力大小为 0.0080 N，方向与电场相反。
The uniform model excludes edge fringing.
均匀场模型排除了边缘效应。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。