Points A and B shown above are in the plane of the page and \(5\,\mathrm{m}\) apart. The points are located in a uniform electric field of magnitude \(1000\,\mathrm{V/m}\) directed toward the bottom of the page. When a proton (of charge \(+e\)) moves from point A to point B, how much work is done on the proton by the electric field?
Correct Answer: \( \boxed{\mathrm{B}} \)
The work done by the electric field is
\(W=qE\Delta d\cos\theta.\)
The proton moves along a path making an angle of \(37^\circ\) with the horizontal, while the electric field is directed vertically downward. Therefore,
\(\theta=90^\circ+37^\circ=127^\circ,\)
or equivalently,
\(W=-qE(5)\cos53^\circ.\)
Using \(\cos53^\circ=0.6\),
\(W=-(e)(1000)(5)(0.6)=-3000\,\mathrm{eV}.\)
The negative sign indicates that the proton moves against the direction of the electric field, so the electric field does negative work.
Therefore, the correct answer is (B).
A helium nucleus (charge \(+2q\) and mass \(4m\)) and a lithium nucleus (charge \(+3q\) and mass \(7m\)) are accelerated through the same electric potential difference, \(V_{0}\).
What is the ratio of their resultant kinetic energies, \( \dfrac{K_{\mathrm{Lithium}}}{K_{\mathrm{Helium}}} \)?
Correct Answer: \( \boxed{\mathrm{D}} \)
When a charged particle is accelerated through a potential difference, the work done by the electric field becomes kinetic energy:
\( K=q\Delta V \)
Since both nuclei are accelerated through the same potential difference \(V_{0}\),
\( K_{\mathrm{Helium}}=(2q)V_{0} \)
\( K_{\mathrm{Lithium}}=(3q)V_{0} \)
Therefore,
\( \dfrac{K_{\mathrm{Lithium}}}{K_{\mathrm{Helium}}}=\dfrac{3qV_{0}}{2qV_{0}}=\dfrac{3}{2} \)
The masses of the nuclei affect their final speeds but do not affect the kinetic energy gained from the same potential difference.
Therefore, the correct answer is (D).
A \(300\,\mathrm{eV}\) electron is aimed midway between two parallel metal plates with a potential difference of \(400\,\mathrm{V}\). The electron is deflected upward and strikes the upper plate as shown.
What would be the kinetic energy of the electron just before striking the upper metal plate?
Correct Answer: \( \boxed{\mathrm{C}} \)
The change in kinetic energy of a charged particle moving through a potential difference is
\( \Delta K=q\Delta V \)
The electron starts midway between the plates, so it moves through only half of the total potential difference:
\( \Delta V=\dfrac{400\,\mathrm{V}}{2}=200\,\mathrm{V} \)
An electron gains \(1\,\mathrm{eV}\) of kinetic energy for every \(1\,\mathrm{V}\) of potential difference through which it moves. Therefore, the additional kinetic energy is
\( \Delta K=200\,\mathrm{eV} \)
Hence, the final kinetic energy is
\( K_f=300\,\mathrm{eV}+200\,\mathrm{eV}=500\,\mathrm{eV} \)
Therefore, the correct answer is (C).
