Key Derivations for 9630-PH05 Physics | 9630-PH05 物理关键公式推导

📚 Key Derivations for 9630-PH05 Physics | 9630-PH05 物理关键公式推导

This article provides step-by-step derivations of essential formulas from the Oxford AQA International A-Level Physics (9630) Unit 5: Fields and their consequences, based on the specimen paper 2016 v2. Mastering these derivations strengthens your understanding of gravitational, electric, and magnetic fields, capacitance, and electromagnetic induction.

本文针对牛津AQA国际A-Level物理（9630）单元5《场及其影响》2016年样卷v2中的关键公式进行逐步推导。掌握这些推导有助于加深你对引力场、电场、磁场、电容及电磁感应的理解。

1. Gravitational Potential Energy Derivation | 引力势能推导

In a radial gravitational field, the force between two point masses M and m separated by distance r is given by Newton’s law. To bring a mass from infinity to a point, work must be done against this force, defining the gravitational potential.

在径向引力场中，两个点质量M和m相距r时的作用力由牛顿定律给出。将一个质量从无穷远处移动到某点需要克服引力做功，从而定义引力势。

Start with the force expression:

从力的表达式出发：

F = GMm / r²

引力 F = GMm / r²

The small work dW done when moving m by an infinitesimal displacement dr towards M (against the field) is negative because the gravitational force and displacement are opposite:

当将m朝着M的方向移动微小位移dr时（逆着场），所做的微功dW为负，因为引力与位移方向相反：

dW = -F dr = -(GMm / r²) dr

Integrate from r = ∞ (where potential energy is defined as zero) to a finite distance r:

从r = ∞（该处势能定义为零）积分到有限距离r：

W = ∫ (∞ → r) -(GMm / r²) dr = GMm [1/r]∞^r = -GMm / r

This work done is the gravitational potential energy U of the mass m. Dividing by m gives the gravitational potential V:

这个功就是质量m的引力势能U。除以m就得到引力势V：

U = -GMm / r , V = -GM / r

引力势能 U = -GMm / r ，引力势 V = -GM / r

2. Electric Potential of a Point Charge | 点电荷的电势

The electric force on a test charge q due to a point charge Q follows Coulomb’s law. By an analogous integration from infinity, we obtain the electric potential energy and the potential V for a point charge.

试探电荷q受到点电荷Q的电场力服从库仑定律。通过类似的从无穷远积分的过程，可以得到电势能和点电荷的电势V。

Coulomb’s law for the force:

库仑力表达式：

F = (1 / 4πε₀) · Qq / r²

To bring the test charge q from infinity to a distance r from Q, the external work dW against the electric force is:

将试探电荷q从无穷远移到距离Q为r处，外力克服电场力所做的微功dW为：

dW = -F dr = -(1 / 4πε₀) · Qq / r² dr

dW = -(1 / 4πε₀) · Qq / r² dr

Integrate from ∞ to r:

从∞到r积分：

W = ∫ (∞ → r) -(1/4πε₀) · Qq/r² dr = (1/4πε₀) · Qq [1/r]∞^r = (1/4πε₀) · Qq / r

W = (1/4πε₀) · Qq / r

Thus the electric potential energy U = (1/4πε₀) · Qq / r, and the electric potential V at distance r from Q is:

因此电势能U = (1/4πε₀) · Qq / r，而距离Q为r处的电势V为：

V = U / q = (1/4πε₀) · Q / r

V = (1/4πε₀) · Q / r

3. Energy Stored in a Capacitor | 电容器储存的能量

When a capacitor is charged, work is done to move charge against the growing potential difference. The total energy stored can be derived by integrating the work over the charging process.

电容器充电时需要克服逐渐升高的电势差做功，储存的总能量可以通过对充电过程积分求出。

At any instant, if a charge q has been transferred, the potential difference across the capacitor is V = q / C. The small work dW needed to add an additional charge dq is:

在任意时刻，若已转移了电荷q，电容器两端的电势差为V = q / C。要再增加微小电荷dq所需的功为：

dW = V dq = (q / C) dq

Integrate from q = 0 to the final charge Q:

从q = 0积分到最终电荷Q：

W = ∫ (0 → Q) (q / C) dq = (1 / 2C) [q²]₀^Q = Q² / (2C)

W = ∫₀^Q (q/C) dq = (1/2C)[q²]₀^Q = Q²/(2C)

This work is stored as electrical potential energy E. Using Q = CV, the energy can be expressed in three common forms:

该功以电势能E的形式储存。利用Q = CV，能量可以表示为三种常见形式：

E = ½ QV = ½ CV² = Q² / (2C)

4. Charged Particle in a Uniform Electric Field | 匀强电场中带电粒子的偏转

When a charged particle moves perpendicularly into a uniform electric field, it experiences a constant force, resulting in parabolic motion similar to a projectile. This derivation gives the vertical deflection and trajectory.

带电粒子垂直进入匀强电场时会受到恒力作用，产生类似于抛体运动的抛物线轨迹。本推导给出竖直偏转位移和轨迹。

Consider a particle of charge q and mass m entering with horizontal velocity vₓ into a field E directed vertically. The electric force gives a constant vertical acceleration:

设电荷量为q、质量为m的粒子以水平速度vₓ进入竖直向下的匀强电场E。电场力提供恒定的竖直加速度：

F = qE → a = F / m = qE / m

F = qE ， a = qE / m

Horizontal motion is uniform, so time spent in the field over a horizontal length L is:

水平方向为匀速运动，因此通过水平距离L的时间为：

t = L / vₓ

In the vertical direction, initial velocity is zero, so the vertical displacement y after time t is:

竖直方向初速度为零，经过时间t后的竖直位移y为：

y = ½ a t² = ½ (qE / m) (L / vₓ)² = (qE L²) / (2 m vₓ²)

y = ½ a t² = (qE L²) / (2 m vₓ²)

If the particle is accelerated through a potential difference Vₐ before entering the field, its kinetic energy gives ½ m vₓ² = qVₐ, so y can also be written as:

若粒子在进入电场前被电势差Vₐ加速，由动能定理有½ m vₓ² = qVₐ，因此y还可表示为：

y = (E L²) / (4 Vₐ)

5. Circular Motion of a Charged Particle in a Magnetic Field | 磁场中带电粒子的圆周运动

A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic force that acts as a centripetal force, causing the particle to move in a circular path. The radius and period can be derived.

带电粒子垂直进入匀强磁场时，受到的洛伦兹力充当向心力，使粒子做匀速圆周运动，可以导出轨道半径和周期。

For a charge q moving with speed v perpendicular to field B, the magnetic force is:

电荷q以速度v垂直于磁场B运动时，磁力大小为：

F = B q v

This force provides the centripetal force required for circular motion of radius r:

该力提供圆周运动所需的向心力：

B q v = m v² / r

Solving for the radius r gives:

解得轨道半径r：

r = m v / (B q)

The period T for one complete revolution is the circumference divided by speed:

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Key Derivations for 9630-PH05 Physics | 9630-PH05 物理关键公式推导