Electromagnetism
GRE Subject Test · GRE Physics Topic 2 4:14 English narration · English + 中文 subtitles burned in
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Transcript
Zero electric potential at a point does not mean zero electric field there: the field depends on how the potential changes, not on its value.
某点电势为零并不意味着该点电场为零：电场取决于电势如何变化，而不是电势的数值。
This topic covers electrostatic potentials, fields and Gauss's law; circuits, magnetism and induction; and Maxwell's equations with electromagnetic waves.
本主题涵盖静电势、电场和高斯定律；电路、磁学和电磁感应；以及麦克斯韦方程组和电磁波。
Electric field is the negative gradient of potential.
电场是电势的负梯度。
A value and a spatial change are different quantities.
一个数值与空间变化是不同的物理量。
Use Gauss’s law with symmetry.
利用高斯定律结合对称性。
A known total flux alone does not give the field at every surface point.
仅已知总通量无法确定表面上每一点的场强。
Kirchhoff’s laws track charge and energy in circuits.
基尔霍夫定律追踪电路中的电荷与能量。
Induced emf follows changing magnetic flux with the Lenz-law sign.
感应电动势遵循磁通量变化，并带有楞次定律的符号。
Maxwell equations link electric and magnetic fields.
麦克斯韦方程组将电场和磁场联系起来。
In vacuum a wave has E/B=c and carries energy.
在真空中，波满足E除以B等于c，并携带能量。
A spherical shell has radius R and uniform charge Q.
一个球壳的半径为R，带有均匀电荷Q。
Choose zero potential at infinity.
选取无穷远处电势为零。
Spherical symmetry lets Gauss's law determine the field outside.
球对称性使得高斯定律能够确定外部电场。
Outside the shell, take a sphere of radius r as the Gaussian surface.
在球壳外，取半径为 r 的球面作为高斯面。
The flux, four pi r squared times E, equals Q over epsilon zero, so E equals Q over four pi epsilon zero r squared: the same field as a point charge.
通量等于 4πr² 乘 E，等于 Q 除以 ε₀，所以 E 等于 Q 除以 4πε₀r²，与点电荷的电场相同。
Inside, the enclosed charge is zero, and symmetry gives E equal to zero.
在球壳内部，包围的电荷为零，由对称性得 E 等于零。
Continuity of the potential at the shell then fixes V everywhere.
再由球壳处电势的连续性，即可确定各处的电势 V。
Inside, V is constant at Q over four pi epsilon zero R. Outside, V equals Q over four pi epsilon zero r, falling off as one over r, while the field falls off as one over r squared.
球壳内，V 恒等于 Q 除以 4πε₀R；球壳外，V 等于 Q 除以 4πε₀r，按 1/r 递减，而电场按 1/r² 递减。
The potential is nonzero and constant inside.
电势在内部非零且为常数。
A zero gradient, rather than a zero value, gives a zero field.
梯度为零，而非数值为零，导致场强为零。
The radial field jumps at the charged surface.
径向场在带电表面处发生跃变。
The handout also shows a capacitor discharging through a resistor.
讲义还给出了电容器通过电阻放电的图像。
The voltage falls exponentially: after one time constant, R C, it has dropped to one over e, about 37 percent of its starting value, and after five time constants it is almost zero.
电压按指数规律下降：经过一个时间常数 RC 后，降到初值的 1/e，约 37%；经过五个时间常数后几乎为零。
Two concentric thin spherical shells have radii R and 2R, and charges Q and negative Q.
两个同心薄球壳的半径分别为 R 和 2R，电荷量分别为 Q 和负 Q。
Derive E and V in all three radial regions, with zero potential at infinity.
推导三个径向区域中的电场强度 E 和电势 V，规定无穷远处电势为零。
Check continuity of V at both shells.
检验两球壳处电势 V 的连续性。
Six marks.
六分。
Let k equal one divided by four pi epsilon zero, and positive field means radially outward.
设 k 等于 1 除以 4πϵ₀，正场表示径向向外。
For r less than R, E equals zero, and V equals k Q over R minus k Q over two R, which simplifies to k Q over two R.
当 r 小于 R 时，E 等于零，V 等于 kQ/R 减去 kQ/2R，即 kQ/2R。
For R less than r less than 2R, E equals k Q over r squared, and V equals k Q over r minus k Q over two R.
当 R 小于 r 小于 2R 时，E 等于 kQ/r²，V 等于 kQ/r 减去 kQ/2R。
For r greater than 2R, both E and V are zero.
当 r 大于 2R 时，E 和 V 均为零。
At R, the two potentials give k Q over two R; at 2R, they give zero.
在 R 处，两个电势给出 kQ/2R；在 2R 处，它们给出零。
Normal E jumps at the surface charges.
法向电场在表面电荷处发生跃变。
Award each region's pair one mark, each continuity check one, and field-jump interpretation one.
每个区域的对应项给一分，每个连续性检查给一分，电场跃变的解释给一分。
Do not use Gauss’s law to claim a uniform field on an arbitrary nonsymmetric surface.
请勿使用高斯定律声称在任意非对称表面上存在均匀场。
flux: The surface integral of a field’s normal component.
通量：场法向分量的面积分。
potential: Potential energy per unit charge.
电势：单位电荷的电势能。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆其含义，然后改变原始传递中的一个参数并检验极限情况。