高级物理学1A|PHYS1131 Physics 1B代写 unsw

This is the second of the two introductory courses in Physics. It is a calculus based course. The course is examined at two levels, with Physics 1A being the lower of the two levels.Electricity and Magnetism: electrostatics, Gauss’s law, electric potential, capacitance and dielectrics, magnetic fields and magnetism, Ampere’s and Biot-Savart law, Faraday’s law, induction and inductance. Physical Optics: light, interference, diffraction, gratings and spectra, polarization. Introductory quantum theory and the wave nature of matter. Introductory solid state and semiconductor physics: simple energy band picture.

这是一份unsw新南威尔斯大学PHYS1131的成功案例

高级物理学1A|PHYS1131 Physics 1B代写 unsw代写


\begin{prob}

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $E=0$ ) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

证明 .

(a) Let the equilibrium of the test charge be stable. If test charge is in equilibrium and it is displaced from its position in any direction, then it experiences a force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards and towards the null point. There is a net inward flux through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at some distance. The mid-point of the joining line of the two charges is the null point. When a test charged is displaced along the line joining charges, it experiences a restoring force. If it is displaced normal to the joining line of charges, then the net force takes it away from the null point. Hence, the charge is not stable because stability of equilibrium requires restoring force in all directions.






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PHYS1131 COURSE NOTES :

What is the force between two small charged sphere having charges of $2 \times$ $10^{-7} \mathrm{C}$ and $3 \times 10^{-7} \mathrm{C}$ placed $30 \mathrm{~cm}$ apart in air?
Solution:
Given:
Charge on the first sphere, $\mathrm{Q}{1}=2 \times 10^{-7} \mathrm{C}$ Charge on the first sphere, $Q{2}=3 \times 10^{-7} \mathrm{C}$
Distance between the spheres is, $r=30 \mathrm{~cm}, \mathrm{r}=0.3 \mathrm{~m}$
The electrostatic force between two charges is given by the formula, $\mathrm{F}=\frac{\mathrm{kQ}{1} \mathrm{Q}{2}}{\mathrm{r}^{2}}$
Where $\mathrm{k}=\frac{1}{4 \pi \varepsilon}=9 \times 10^{9}$
$$
\mathrm{F}=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} \mathrm{~N}
$$
Hence, the force between the spheres is $6 \times 10^{-3} \mathrm{~N}$. Charges are of the same polarity so force between them will be repulsive in nature.




















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