201. A particle of charge \(+2e\) is accelerated from rest through a potential difference of \(50\,\text{V}\). What kinetic energy does it gain?
ⓐ. \(25\,\text{eV}\)
ⓑ. \(50\,\text{eV}\)
ⓒ. \(100\,\text{eV}\)
ⓓ. \(200\,\text{eV}\)
Correct Answer: \(100\,\text{eV}\)
Explanation: \(\textbf{Charge:}\) The particle has charge \(+2e\).
\(\textbf{Potential difference:}\) \(\Delta V=50\,\text{V}\).
\(\textbf{Energy gained:}\) A charge \(q\) gains energy \(q\Delta V\) when accelerated through a potential difference.
\(\textbf{In electron-volt form:}\) A charge \(e\) through \(1\,\text{V}\) corresponds to \(1\,\text{eV}\).
\(\textbf{Applying charge factor:}\) A charge \(2e\) through \(50\,\text{V}\) gains \(2\times50\,\text{eV}\).
\(\textbf{Calculation:}\) \(K=100\,\text{eV}\).
\(\textbf{Joule check:}\) \(100\,\text{eV}=1.6\times10^{-17}\,\text{J}\).
\(\textbf{Final result:}\) The kinetic energy gained is \(100\,\text{eV}\).
202. An alpha particle has charge \(+2e\). If it gains \(6.4\times10^{-17}\,\text{J}\) of kinetic energy, through what potential difference was it accelerated?
ⓐ. \(100\,\text{V}\)
ⓑ. \(150\,\text{V}\)
ⓒ. \(250\,\text{V}\)
ⓓ. \(200\,\text{V}\)
Correct Answer: \(200\,\text{V}\)
Explanation: \(\textbf{Given energy:}\) \(K=6.4\times10^{-17}\,\text{J}\).
\(\textbf{Charge:}\) For an alpha particle, \(q=+2e=2(1.6\times10^{-19})=3.2\times10^{-19}\,\text{C}\).
\(\textbf{Required:}\) Accelerating potential difference \(\Delta V\).
\(\textbf{Energy relation:}\) \(K=q\Delta V\).
\(\textbf{Solving:}\) \(\Delta V=\frac{K}{q}\).
\(\textbf{Substitution:}\) \(\Delta V=\frac{6.4\times10^{-17}}{3.2\times10^{-19}}\,\text{V}\).
\(\textbf{Simplification:}\) \(\frac{6.4}{3.2}=2.0\) and \(10^{-17}/10^{-19}=10^2\).
\(\textbf{Calculation:}\) \(\Delta V=2.0\times10^2\,\text{V}=200\,\text{V}\).
\(\textbf{Final result:}\) The required potential difference is \(200\,\text{V}\).
203. An electron is accelerated from rest through \(150\,\text{V}\). What is its kinetic energy in joules?
ⓐ. \(1.5\times10^{-19}\,\text{J}\)
ⓑ. \(2.4\times10^{-18}\,\text{J}\)
ⓒ. \(1.5\times10^{-17}\,\text{J}\)
ⓓ. \(2.4\times10^{-17}\,\text{J}\)
Correct Answer: \(2.4\times10^{-17}\,\text{J}\)
Explanation: \(\textbf{Potential difference:}\) The electron is accelerated through \(150\,\text{V}\).
\(\textbf{Energy in electron volts:}\) A particle of charge magnitude \(e\) gains \(150\,\text{eV}\).
\(\textbf{Conversion factor:}\) \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\).
\(\textbf{Conversion to joules:}\) \(K=150(1.6\times10^{-19})\,\text{J}\).
\(\textbf{Calculation:}\) \(150\times1.6=240\).
\(\textbf{Power of ten:}\) \(240\times10^{-19}\,\text{J}=2.4\times10^{-17}\,\text{J}\).
\(\textbf{Sign note:}\) Kinetic energy gained is positive even though the electron charge is negative.
\(\textbf{Final result:}\) The kinetic energy is \(2.4\times10^{-17}\,\text{J}\).
204. A charged particle gains \(960\,\text{eV}\) of kinetic energy when accelerated through \(240\,\text{V}\). What is the magnitude of its charge?
ⓐ. \(e\)
ⓑ. \(4e\)
ⓒ. \(2e\)
ⓓ. \(8e\)
Correct Answer: \(4e\)
Explanation: \(\textbf{Given energy:}\) \(K=960\,\text{eV}\).
\(\textbf{Potential difference:}\) \(\Delta V=240\,\text{V}\).
\(\textbf{Energy relation in \(\text{eV}\):}\) A charge \(ne\) accelerated through \(\Delta V\) gains \(n\Delta V\,\text{eV}\).
\(\textbf{Set up equation:}\) \(960=n(240)\).
\(\textbf{Solving:}\) \(n=\frac{960}{240}=4\).
\(\textbf{Charge magnitude:}\) \(|q|=4e\).
\(\textbf{Reasonableness check:}\) A singly charged particle through \(240\,\text{V}\) would gain \(240\,\text{eV}\), so \(960\,\text{eV}\) requires four times the elementary charge.
\(\textbf{Final result:}\) The magnitude of charge is \(4e\).
205. Which statement is correct for an electron accelerated freely from lower electric potential to higher electric potential?
ⓐ. Its kinetic energy increases
ⓑ. Its kinetic energy decreases
ⓒ. Its charge becomes positive
ⓓ. Its potential energy increases
Correct Answer: Its kinetic energy increases
Explanation: An electron has negative charge. Its potential energy is \(U=qV=-eV\). When an electron moves to a higher electric potential, \(V\) increases, so \(U\) decreases because of the negative sign of \(q\). If the electron moves freely under the electric field, the decrease in potential energy appears as an increase in kinetic energy. This is why electrons gain kinetic energy when accelerated through a positive potential difference in the direction appropriate for their negative charge. The charge of the electron does not change during the process. The key idea is that a negative charge loses potential energy when it moves to higher potential. Therefore, its kinetic energy increases.
206. A particle of charge \(-2e\) moves through a potential difference \(V_B-V_A=+75\,\text{V}\). What is its change in potential energy?
ⓐ. \(+75\,\text{eV}\)
ⓑ. \(-75\,\text{eV}\)
ⓒ. \(+150\,\text{eV}\)
ⓓ. \(-150\,\text{eV}\)
Correct Answer: \(-150\,\text{eV}\)
Explanation: \(\textbf{Charge:}\) \(q=-2e\).
\(\textbf{Potential difference:}\) \(\Delta V=V_B-V_A=+75\,\text{V}\).
\(\textbf{Potential-energy relation:}\) \(\Delta U=q\Delta V\).
\(\textbf{Substitution:}\) \(\Delta U=(-2e)(75\,\text{V})\).
\(\textbf{Electron-volt conversion:}\) \(e\cdot1\,\text{V}=1\,\text{eV}\).
\(\textbf{Calculation:}\) \(\Delta U=-150\,\text{eV}\).
\(\textbf{Sign meaning:}\) The negative charge loses potential energy when moving through a positive potential difference.
\(\textbf{Final result:}\) The change in potential energy is \(-150\,\text{eV}\).
207. A proton gains \(3.2\times10^{-16}\,\text{J}\) of kinetic energy. Through what potential difference was it accelerated, using \(e=1.6\times10^{-19}\,\text{C}\)?
ⓐ. \(1.0\times10^3\,\text{V}\)
ⓑ. \(2.0\times10^3\,\text{V}\)
ⓒ. \(3.2\times10^3\,\text{V}\)
ⓓ. \(5.0\times10^3\,\text{V}\)
Correct Answer: \(2.0\times10^3\,\text{V}\)
Explanation: \(\textbf{Given kinetic energy:}\) \(K=3.2\times10^{-16}\,\text{J}\).
\(\textbf{Charge of proton:}\) \(q=+e=1.6\times10^{-19}\,\text{C}\).
\(\textbf{Energy relation:}\) \(K=q\Delta V\).
\(\textbf{Solving for potential difference:}\) \(\Delta V=\frac{K}{q}\).
\(\textbf{Substitution:}\) \(\Delta V=\frac{3.2\times10^{-16}}{1.6\times10^{-19}}\,\text{V}\).
\(\textbf{Simplification:}\) \(\frac{3.2}{1.6}=2.0\) and \(10^{-16}/10^{-19}=10^3\).
\(\textbf{Calculation:}\) \(\Delta V=2.0\times10^3\,\text{V}\).
\(\textbf{Final result:}\) The proton was accelerated through \(2.0\times10^3\,\text{V}\).
208. A charged particle of charge \(+3e\) is accelerated from rest through \(120\,\text{V}\). What kinetic energy does it gain in joules?
ⓐ. \(1.92\times10^{-17}\,\text{J}\)
ⓑ. \(3.60\times10^{-17}\,\text{J}\)
ⓒ. \(5.76\times10^{-17}\,\text{J}\)
ⓓ. \(1.20\times10^{-17}\,\text{J}\)
Correct Answer: \(5.76\times10^{-17}\,\text{J}\)
Explanation: \(\textbf{Charge:}\) \(q=+3e=3(1.6\times10^{-19})\,\text{C}=4.8\times10^{-19}\,\text{C}\).
\(\textbf{Potential difference:}\) \(\Delta V=120\,\text{V}\).
\(\textbf{Energy relation:}\) \(K=q\Delta V\).
\(\textbf{Substitution:}\) \(K=(4.8\times10^{-19})(120)\,\text{J}\).
\(\textbf{Calculation:}\) \(4.8\times120=576\).
\(\textbf{Power of ten:}\) \(K=576\times10^{-19}\,\text{J}\).
\(\textbf{Scientific notation:}\) \(K=5.76\times10^{-17}\,\text{J}\).
\(\textbf{Electron-volt check:}\) The same result is \(3\times120=360\,\text{eV}\), and \(360(1.6\times10^{-19})=5.76\times10^{-17}\,\text{J}\).
\(\textbf{Final result:}\) The kinetic energy gained is \(5.76\times10^{-17}\,\text{J}\).
209. A particle gains \(2.5\,\text{keV}\) of kinetic energy. What is this energy in joules, using \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\)?
ⓐ. \(4.0\times10^{-16}\,\text{J}\)
ⓑ. \(2.5\times10^{-16}\,\text{J}\)
ⓒ. \(1.6\times10^{-16}\,\text{J}\)
ⓓ. \(4.0\times10^{-19}\,\text{J}\)
Correct Answer: \(4.0\times10^{-16}\,\text{J}\)
Explanation: \(\textbf{Given energy:}\) \(2.5\,\text{keV}=2.5\times10^3\,\text{eV}\).
\(\textbf{Conversion factor:}\) \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\).
\(\textbf{Required:}\) Energy in joules.
\(\textbf{Conversion relation:}\) \(E(\text{J})=E(\text{eV})(1.6\times10^{-19})\).
\(\textbf{Substitution:}\) \(E=(2.5\times10^3)(1.6\times10^{-19})\,\text{J}\).
\(\textbf{Number multiplication:}\) \(2.5\times1.6=4.0\).
\(\textbf{Power of ten:}\) \(10^3\times10^{-19}=10^{-16}\).
\(\textbf{Final result:}\) \(E=4.0\times10^{-16}\,\text{J}\).
210. An electron is accelerated from rest through \(100\,\text{V}\). Using \(e=1.6\times10^{-19}\,\text{C}\) and \(m_e=9.1\times10^{-31}\,\text{kg}\), what is its approximate speed?
ⓐ. \(2.0\times10^6\,\text{m s}^{-1}\)
ⓑ. \(4.2\times10^6\,\text{m s}^{-1}\)
ⓒ. \(5.9\times10^6\,\text{m s}^{-1}\)
ⓓ. \(8.4\times10^6\,\text{m s}^{-1}\)
Correct Answer: \(5.9\times10^6\,\text{m s}^{-1}\)
Explanation: \(\textbf{Given:}\) \(\Delta V=100\,\text{V}\), \(e=1.6\times10^{-19}\,\text{C}\), and \(m_e=9.1\times10^{-31}\,\text{kg}\).
\(\textbf{Energy gained:}\) The electron starts from rest, so electric work becomes kinetic energy.
\(\textbf{Energy relation:}\) \(e\Delta V=\frac{1}{2}m_ev^2\).
\(\textbf{Solving for speed:}\) \(v=\sqrt{\frac{2e\Delta V}{m_e}}\).
\(\textbf{Substitution:}\) \(v=\sqrt{\frac{2(1.6\times10^{-19})(100)}{9.1\times10^{-31}}}\).
\(\textbf{Numerator:}\) \(2(1.6\times10^{-19})(100)=3.2\times10^{-17}\).
\(\textbf{Division:}\) \(\frac{3.2\times10^{-17}}{9.1\times10^{-31}}\approx3.52\times10^{13}\).
\(\textbf{Square root:}\) \(v\approx\sqrt{3.52\times10^{13}}\approx5.9\times10^6\,\text{m s}^{-1}\).
\(\textbf{Final result:}\) The electron speed is approximately \(5.9\times10^6\,\text{m s}^{-1}\).
211. A singly charged ion gains \(1.2\,\text{keV}\) of kinetic energy from rest. Through what potential difference was it accelerated?
ⓐ. \(120\,\text{V}\)
ⓑ. \(1200\,\text{V}\)
ⓒ. \(2400\,\text{V}\)
ⓓ. \(600\,\text{V}\)
Correct Answer: \(1200\,\text{V}\)
Explanation: \(\textbf{Given energy:}\) \(K=1.2\,\text{keV}=1200\,\text{eV}\).
\(\textbf{Charge condition:}\) A singly charged ion has charge magnitude \(e\).
\(\textbf{Electron-volt meaning:}\) A charge of magnitude \(e\) gains \(1\,\text{eV}\) through \(1\,\text{V}\).
\(\textbf{Energy-potential relation:}\) \(K=e\Delta V\).
\(\textbf{In electron-volt form:}\) \(K(\text{eV})=\Delta V(\text{V})\) for a singly charged particle.
\(\textbf{Substitution:}\) \(1200\,\text{eV}\) corresponds to \(1200\,\text{V}\).
\(\textbf{Final result:}\) The ion was accelerated through \(1200\,\text{V}\).
212. Which statement is accurate about using \(\text{eV}\) in electrostatic energy calculations?
ⓐ. \(\text{eV}\) is used only for electric field and never for energy
ⓑ. \(\text{eV}\) is a potential unit equal to \(1.6\times10^{-19}\,\text{V}\)
ⓒ. \(\text{eV}\) is a charge unit equal to \(1.6\times10^{-19}\,\text{C}\)
ⓓ. \(\text{eV}\) is convenient for very small energies of charged particles
Correct Answer: \(\text{eV}\) is convenient for very small energies of charged particles
Explanation: The electron volt is a unit of energy. It is especially useful when dealing with microscopic particles such as electrons, protons, ions, and atoms. Energies at that scale are often very small when written in joules. Since \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\), using \(\text{eV}\) gives simpler numerical values. The unit \(\text{eV}\) should not be confused with \(\text{V}\), which measures potential difference. It is also not a unit of charge; the elementary charge is \(e=1.6\times10^{-19}\,\text{C}\). Thus \(\text{eV}\) is convenient because it represents small energies in a compact form.
213. An electron moves from a point at \(-30\,\text{V}\) to a point at \(+70\,\text{V}\). What is the change in its potential energy?
ⓐ. \(-100\,\text{eV}\)
ⓑ. \(+100\,\text{eV}\)
ⓒ. \(-40\,\text{eV}\)
ⓓ. \(+40\,\text{eV}\)
Correct Answer: \(-100\,\text{eV}\)
Explanation: \(\textbf{Charge of electron:}\) \(q=-e\).
\(\textbf{Initial potential:}\) \(V_A=-30\,\text{V}\).
\(\textbf{Final potential:}\) \(V_B=+70\,\text{V}\).
\(\textbf{Potential difference:}\) \(V_B-V_A=70-(-30)=100\,\text{V}\).
\(\textbf{Energy relation:}\) \(\Delta U=q(V_B-V_A)\).
\(\textbf{Substitution:}\) \(\Delta U=(-e)(100\,\text{V})\).
\(\textbf{Electron-volt conversion:}\) \(e\cdot1\,\text{V}=1\,\text{eV}\).
\(\textbf{Final result:}\) \(\Delta U=-100\,\text{eV}\).
214. A particle of charge \(-3e\) moves from \(A\) to \(B\), where \(V_B-V_A=-40\,\text{V}\). What is the change in its electrostatic potential energy?
ⓐ. \(-120\,\text{eV}\)
ⓑ. \(-40\,\text{eV}\)
ⓒ. \(+40\,\text{eV}\)
ⓓ. \(+120\,\text{eV}\)
Correct Answer: \(+120\,\text{eV}\)
Explanation: \(\textbf{Charge:}\) \(q=-3e\).
\(\textbf{Potential difference:}\) \(\Delta V=V_B-V_A=-40\,\text{V}\).
\(\textbf{Potential-energy relation:}\) \(\Delta U=q\Delta V\).
\(\textbf{Substitution:}\) \(\Delta U=(-3e)(-40\,\text{V})\).
\(\textbf{Sign calculation:}\) The product of two negative factors is positive.
\(\textbf{Magnitude:}\) \(3\times40=120\).
\(\textbf{Electron-volt conversion:}\) \(e\cdot1\,\text{V}=1\,\text{eV}\).
\(\textbf{Final result:}\) \(\Delta U=+120\,\text{eV}\).
215. An electric dipole consists of charges \(+q\) and \(-q\). If a point \(P\) is at distances \(r_+\) from \(+q\) and \(r_-\) from \(-q\), which expression gives the exact potential at \(P\)?
ⓐ. \(V=kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\)
ⓑ. \(V=kq\left(\frac{1}{r_+^2}-\frac{1}{r_-^2}\right)\)
ⓒ. \(V=kq\left(\frac{1}{r_+}+\frac{1}{r_-}\right)\)
ⓓ. \(V=kq\left(\frac{r_+}{r_-}-\frac{r_-}{r_+}\right)\)
Correct Answer: \(V=kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\)
Explanation: \(\textbf{Dipole charges:}\) The charges are \(+q\) and \(-q\).
\(\textbf{Potential due to \(+q\):}\) \(V_+=\frac{kq}{r_+}\).
\(\textbf{Potential due to \(-q\):}\) \(V_-=-\frac{kq}{r_-}\).
\(\textbf{Scalar addition:}\) Electric potentials add algebraically.
\(\textbf{Total potential:}\) \(V=V_++V_-\).
\(\textbf{Substitution:}\) \(V=\frac{kq}{r_+}-\frac{kq}{r_-}\).
\(\textbf{Factor form:}\) \(V=kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\).
\(\textbf{Final result:}\) The exact potential is \(V=kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\).
216. A point \(P\) is closer to the positive charge of a dipole than to the negative charge. Which sign is expected for the electric potential at \(P\)?
ⓐ. Always zero
ⓑ. Positive
ⓒ. Negative
ⓓ. Directionless and signless
Correct Answer: Positive
Explanation: The potential of a dipole at a point is the algebraic sum of the potentials due to \(+q\) and \(-q\). If \(P\) is closer to \(+q\), then the positive contribution \(\frac{kq}{r_+}\) has a larger magnitude than the negative contribution \(-\frac{kq}{r_-}\). Since \(r_+\frac{1}{r_-}\). Therefore, \(kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\) is positive. The potential has no direction, but it can have a sign. The sign tells whether the algebraic scalar sum is above or below the chosen zero reference. Hence the potential at \(P\) is positive.
217. A point \(P\) is at equal distances from the charges \(+q\) and \(-q\) of an electric dipole. What is the potential at \(P\)?
ⓐ. \(\frac{2kq}{r}\)
ⓑ. \(-\frac{2kq}{r}\)
ⓒ. \(0\)
ⓓ. \(\frac{kq}{r^2}\)
Correct Answer: \(0\)
Explanation: \(\textbf{Distance condition:}\) \(P\) is at the same distance \(r\) from \(+q\) and \(-q\).
\(\textbf{Potential due to positive charge:}\) \(V_+=\frac{kq}{r}\).
\(\textbf{Potential due to negative charge:}\) \(V_-=-\frac{kq}{r}\).
\(\textbf{Algebraic addition:}\) \(V=V_++V_-\).
\(\textbf{Substitution:}\) \(V=\frac{kq}{r}-\frac{kq}{r}\).
\(\textbf{Result:}\) \(V=0\).
\(\textbf{Important distinction:}\) Zero potential does not necessarily mean zero electric field, because electric field is vectorial.
\(\textbf{Final result:}\) The potential at \(P\) is \(0\).
218. For an electric dipole, a point \(P\) has \(r_+=0.20\,\text{m}\) and \(r_-=0.30\,\text{m}\), where \(r_+\) and \(r_-\) are distances from \(+q\) and \(-q\). If \(q=2.0\,\mu\text{C}\) and \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\), what is the potential at \(P\)?
ⓐ. \(+3.0\times10^4\,\text{V}\)
ⓑ. \(+1.5\times10^4\,\text{V}\)
ⓒ. \(-1.5\times10^4\,\text{V}\)
ⓓ. \(-3.0\times10^4\,\text{V}\)
Correct Answer: \(+3.0\times10^4\,\text{V}\)
Explanation: \(\textbf{Given:}\) \(q=2.0\times10^{-6}\,\text{C}\), \(r_+=0.20\,\text{m}\), \(r_-=0.30\,\text{m}\), and \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
\(\textbf{Potential formula:}\) \(V=kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\).
\(\textbf{Distance reciprocals:}\) \(\frac{1}{0.20}=5.0\,\text{m}^{-1}\) and \(\frac{1}{0.30}\approx3.33\,\text{m}^{-1}\).
\(\textbf{Difference:}\) \(5.0-3.33\approx1.67\,\text{m}^{-1}\).
\(\textbf{Coefficient:}\) \(kq=(9.0\times10^9)(2.0\times10^{-6})=1.8\times10^4\).
\(\textbf{Substitution:}\) \(V=(1.8\times10^4)(1.67)\,\text{V}\).
\(\textbf{Calculation:}\) \(V\approx3.0\times10^4\,\text{V}\).
\(\textbf{Final result:}\) The potential is \(+3.0\times10^4\,\text{V}\).
219. A point \(P\) is closer to the negative charge of an electric dipole than to the positive charge. Which statement is correct?
ⓐ. The potential must be positive because dipoles contain \(+q\)
ⓑ. The potential must be zero because charges are equal in magnitude
ⓒ. The potential must be infinite because opposite charges are present
ⓓ. The potential is negative due to the larger negative contribution
Correct Answer: The potential is negative due to the larger negative contribution
Explanation: The potential at \(P\) is found by adding the scalar contributions from \(+q\) and \(-q\). If \(P\) is closer to the negative charge, the magnitude of the negative contribution is larger. In the expression \(V=kq\left(\frac{1}{r_+}-\frac{1}{r_-}\right)\), being closer to \(-q\) means \(r_-\frac{1}{r_+}\), making the bracket negative. The equal magnitudes of the charges do not guarantee cancellation unless the distances are equal. The potential remains finite at ordinary points away from the charges. Hence the potential is negative.
220. Which statement best describes why dipole potential may vanish at some points even though charges are present?
ⓐ. The electric field of a dipole is always zero wherever potential is zero
ⓑ. The scalar potentials of \(+q\) and \(-q\) can cancel algebraically
ⓒ. The positive charge stops producing potential at large distances
ⓓ. The negative charge produces only electric field and no potential
Correct Answer: The scalar potentials of \(+q\) and \(-q\) can cancel algebraically
Explanation: Electric potential is a scalar quantity, so contributions from different charges are added with signs. In a dipole, the positive charge contributes positive potential and the negative charge contributes negative potential. At points where the two contributions have equal magnitudes, their algebraic sum becomes zero. This can happen even though both charges are still present and still produce electric fields. Electric field is vectorial, so its cancellation requires a different condition involving directions. A negative charge produces potential just as a positive charge does, but with opposite sign. Therefore, zero dipole potential is caused by scalar cancellation, not by absence of charge effect.
