Conservation of Electric Energy - College Physics

A conducting sphere of radius $R$ carries total charge $Q$ . What is the electric potential energy of this charge distribution?
Answer: $\frac{1}{2} \cdot k\frac{Q^2}{R}$ When you encounter questions about electric potential energy of charge distributions, you're dealing with the self-energy of the system - the energy required to assemble the charges from infinity. For a conducting sphere, all charge $Q$ resides on the surface at radius $R$ , and the electric potential at the surface is $V = k\frac{Q}{R}$ . To find the self-energy, imagine building up the charge gradually. When you've already placed charge $q$ on the sphere, the potential is $V = k\frac{q}{R}$ , and adding a small additional charge $dq$ requires work $dW = V \cdot dq = k\frac{q}{R}dq$ . The total energy is: $U = \int_0^Q k\frac{q}{R}dq = \frac{k}{R}\int_0^Q q \, dq = \frac{k}{R} \cdot \frac{Q^2}{2} = \frac{1}{2}k\frac{Q^2}{R}$ This confirms answer A is correct. Answer B ( $k\frac{Q^2}{R}$ ) omits the crucial factor of $\frac{1}{2}$ that comes from the integration - a common mistake when students forget that potential energy involves building up charge gradually. Answer C ( $\frac{1}{4\pi\epsilon_0} \cdot \frac{Q^2}{2R}$ ) has the wrong denominator factor. Since $k = \frac{1}{4\pi\epsilon_0}$ , this would give $\frac{1}{2}k\frac{Q^2}{2R}$ , which is off by a factor of 2 in the denominator. Answer D ( $\frac{1}{8\pi\epsilon_0} \cdot \frac{Q^2}{R}$ ) represents $\frac{1}{2}k\frac{Q^2}{2R}$ , combining both errors from options B and C. Study tip: Always remember the $\frac{1}{2}$ factor in electrostatic self-energy problems - it comes from gradually assembling the charge distribution, not placing it all at once.
A parallel-plate capacitor is charged and then disconnected from the battery. A dielectric material with dielectric constant $\kappa = 3$ is then inserted between the plates. How does the energy stored in the capacitor change?
Answer: The energy decreases by a factor of 3 When you encounter capacitor problems involving dielectrics, pay close attention to whether the capacitor remains connected to or is disconnected from the battery, as this determines which quantities stay constant. Since the capacitor is disconnected from the battery before the dielectric is inserted, the charge $Q$ on the plates remains constant throughout the process. The energy stored in a capacitor can be expressed as $U = \frac{Q^2}{2C}$ , making it clear how energy depends on capacitance when charge is fixed. When the dielectric with $\kappa = 3$ is inserted, the capacitance increases by the factor $\kappa$ : $C_{new} = \kappa C = 3C$ . Since $U = \frac{Q^2}{2C}$ and $Q$ is constant, the new energy becomes $U_{new} = \frac{Q^2}{2(3C)} = \frac{1}{3} \cdot \frac{Q^2}{2C} = \frac{U_{original}}{3}$ . The energy decreases by a factor of 3, confirming answer C. Option A incorrectly assumes energy increases by $\kappa$ , which would happen if voltage were constant (battery connected). Option B suggests energy increases by $\kappa^2$ , confusing this with scenarios involving electric field relationships. Option D incorrectly applies the $\kappa^2$ factor as a decrease, mixing up the physics of different capacitor configurations. Key strategy: Always identify whether charge or voltage remains constant in capacitor problems. When disconnected from the battery, charge is constant, and inserting a dielectric always decreases the stored energy by the factor $\kappa$ .
An alpha particle (charge $+2e$ ) approaches a stationary gold nucleus (charge $+79e$ ) head-on with initial kinetic energy $K_0$ . At the distance of closest approach, what is the ratio of electric potential energy to the initial kinetic energy?
Answer: $1$ because all initial kinetic energy converts to potential energy at closest approach At the distance of closest approach, the alpha particle momentarily stops, so all its initial kinetic energy converts to electric potential energy by conservation of energy: $K_0 + U_i = K_f + U_f$ . Since $K_f = 0$ and $U_i = 0$ (infinite separation), we have $K_0 = U_f$ . Therefore, the ratio $\frac{U_f}{K_0} = 1$ . Choice A incorrectly relates the charge ratio to energy ratio. Choice C doubles the effect incorrectly. Choice D inverts the charge ratio inappropriately. The key insight is that energy conservation determines the relationship, not the specific charge values.
A charged capacitor with capacitance $C$ and initial voltage $V_0$ is connected through a resistor to an identical uncharged capacitor. After the system reaches equilibrium, what fraction of the initial electric potential energy is dissipated as heat in the resistor?
Answer: $\frac{1}{2}$ because half the energy is lost during the charge redistribution process Initial energy: $U_i = \frac{1}{2}CV_0^2$ . At equilibrium, charge is shared equally, so each capacitor has charge $Q/2$ where $Q = CV_0$ . The voltage across each becomes $V_f = \frac{Q/2}{C} = \frac{V_0}{2}$ . Final energy: $U_f = 2 \times \frac{1}{2}C\left(\frac{V_0}{2}\right)^2 = \frac{CV_0^2}{4}$ . Energy dissipated: $U_i - U_f = \frac{1}{2}CV_0^2 - \frac{1}{4}CV_0^2 = \frac{1}{4}CV_0^2$ . Fraction dissipated: $\frac{1/4 \cdot CV_0^2}{1/2 \cdot CV_0^2} = \frac{1}{2}$ . Choice A confuses final energy distribution with energy loss. Choice B uses incorrect voltage calculation. Choice D overestimates the dissipation.
Two identical charges, each of magnitude $+q$ , are initially separated by distance $d$ . One charge is held fixed while the other is moved to a distance $2d$ away. If the initial electric potential energy of the system was $U_0$ , what is the final potential energy?
Answer: $\frac{U_0}{2}$ This question tests your understanding of electric potential energy and how it depends on the distance between charges. When dealing with point charges, remember that potential energy follows an inverse relationship with separation distance. The electric potential energy between two point charges is given by $U = k\frac{q_1q_2}{r}$ , where $k$ is Coulomb's constant and $r$ is the separation distance. Initially, with both charges at distance $d$ , we have $U_0 = k\frac{q^2}{d}$ . When one charge moves to distance $2d$ , the final potential energy becomes $U_f = k\frac{q^2}{2d}$ . To find the relationship, divide the final energy by the initial: $\frac{U_f}{U_0} = \frac{k\frac{q^2}{2d}}{k\frac{q^2}{d}} = \frac{1}{2}$ . Therefore, $U_f = \frac{U_0}{2}$ , making answer B correct. Let's examine why the other options are wrong. Choice A ( $\frac{U_0}{4}$ ) incorrectly applies an inverse-square relationship, confusing potential energy with electric field strength. Choice C ( $2U_0$ ) gets the direction backwards, suggesting energy increases with distance for like charges. Choice D ( $4U_0$ ) compounds this error by applying a square relationship in the wrong direction. Remember this key pattern: electric potential energy between point charges is inversely proportional to their separation distance. When distance doubles, potential energy halves. This is different from gravitational or electric field strength, which follow inverse-square laws. Always check whether the question asks for energy ( $\propto \frac{1}{r}$ ) or force/field ( $\propto \frac{1}{r^2}$ ).
A charged particle moves in a region where both gravitational and electric forces are present. The particle moves from point P to point Q along two different paths. If only conservative forces act on the particle, which statement about the change in total potential energy is correct?
Answer: The change in total potential energy is the same for both paths and equals the negative of the change in kinetic energy When analyzing motion under conservative forces like gravity and electric fields, you need to understand the fundamental property that defines conservative forces: the work they do depends only on the initial and final positions, not the path taken. For conservative forces, the change in potential energy between two points is always the same regardless of path. Since both gravitational and electric forces are conservative, the total potential energy change (gravitational + electric) from P to Q must be identical for both paths. Additionally, by the work-energy theorem, this change equals the negative of the work done by conservative forces, which in turn equals the change in kinetic energy. Therefore, the change in total potential energy is the same for both paths and equals the change in kinetic energy. Let's examine why the other options fail: Option A incorrectly suggests path dependence, which would only be true for non-conservative forces. Option B correctly states that the potential energy change is path-independent and relates to work done by forces, but it's incomplete since it doesn't mention the relationship to kinetic energy change. Option C contains a serious misconception—conservative forces definitely do work; they just do the same amount of work regardless of path. Option D correctly identifies that potential energy change equals kinetic energy change but fails to mention the crucial path-independence property. Remember: For conservative force problems, always think about two key properties—path independence and energy conservation. These principles will guide you to recognize that multiple relationships must hold simultaneously.
Two parallel plates are separated by distance $d$ with a potential difference $V$ between them. A charged particle starts from rest at the negative plate and reaches the positive plate. If the plate separation is increased to $2d$ while keeping the same potential difference $V$ , how does the final kinetic energy of the particle compare to the original case?
Answer: The final kinetic energy is the same When dealing with charged particles moving between parallel plates, you need to connect electric fields, forces, and energy concepts. The key insight is understanding what determines the particle's final kinetic energy. The particle gains kinetic energy equal to the work done by the electric field: $KE = qV$ , where $q$ is the particle's charge and $V$ is the potential difference. This comes from the work-energy theorem - the work done moving a charge through a potential difference $V$ is simply $W = qV$ , regardless of the path taken. Since the potential difference $V$ remains constant in both scenarios, the final kinetic energy stays the same. The particle starts from rest and ends with $KE = qV$ in both cases, making answer C correct. Now for the incorrect options: Answer A (four times larger) likely stems from incorrectly thinking that since the electric field $E = V/d$ becomes half as strong, and the distance doubles, some quadratic relationship emerges. Answer B (two times larger) might come from thinking the doubled distance means doubled work, forgetting that the field strength also changes. Answer D (half as large) probably results from focusing only on the weaker electric field ( $E = V/2d$ ) without considering that the particle travels twice the distance. Remember this key principle: when a charged particle moves through a fixed potential difference, its kinetic energy change depends only on its charge and that potential difference - not on the distance traveled or field strength. This makes potential energy particularly powerful for solving electrostatics problems.
In a parallel-plate capacitor, the electric potential varies linearly from $0$ V at the negative plate to $120$ V at the positive plate. An electron moves from the positive plate to the negative plate. What happens to the total mechanical energy of the electron if we ignore external forces?
Answer: The total mechanical energy remains constant by conservation of energy When analyzing motion in electric fields, you need to distinguish between conservative and non-conservative forces. The electric field in a capacitor is conservative, meaning the total mechanical energy (kinetic + potential) of a charged particle remains constant when only electric forces act. As the electron moves from the positive plate (120 V) to the negative plate (0 V), it moves from lower electric potential to higher electric potential. For a negatively charged particle, this means moving from higher electric potential energy to lower electric potential energy. The decrease in electric potential energy equals the increase in kinetic energy, so the electron speeds up during its journey. However, the total mechanical energy stays constant because energy is simply converting from one form (electric potential) to another (kinetic). This is exactly what conservation of energy predicts for conservative force fields. Answer A incorrectly suggests the total mechanical energy increases. While the electric field does positive work on the electron (increasing its kinetic energy), the simultaneous decrease in electric potential energy keeps the total constant. Answer B makes the opposite error, suggesting energy decreases due to work against the field, but the field actually does positive work on the electron in this direction. Answer D incorrectly claims the total mechanical energy becomes zero, confusing the electric potential (which is zero at the negative plate) with mechanical energy. Remember: in conservative force problems, always check whether you're asked about total energy (which stays constant) versus individual energy components (which can change).
An electron gun accelerates electrons from rest through a potential difference. If the potential difference is doubled, by what factor does the final kinetic energy of the electrons change, assuming non-relativistic speeds?
Answer: The kinetic energy increases by a factor of $2$ When you encounter electron gun problems, you're dealing with the relationship between electric potential energy and kinetic energy. The key principle is energy conservation: as an electron accelerates through a potential difference, its electric potential energy converts to kinetic energy. For an electron starting from rest and accelerated through potential $V$ , the work done equals the kinetic energy gained: $W = qV = KE$ . Since the electron's charge is constant, $KE = eV$ , where $e$ is the elementary charge. This shows that kinetic energy is directly proportional to the potential difference. When you double the potential difference ( $2V$ ), the final kinetic energy becomes $KE_{final} = e(2V) = 2eV = 2KE_{initial}$ . The kinetic energy increases by exactly a factor of 2, making answer B correct. Let's examine why the other options fail: Answer A (factor of $\sqrt{2}$ ) incorrectly assumes kinetic energy relates to the square root of potential, perhaps confusing this with how velocity scales (since $v \propto \sqrt{V}$ ). Answer C (factor of 4) wrongly suggests kinetic energy depends on $V^2$ , which would happen if you mistakenly squared the relationship. Answer D claims the kinetic energy decreases, which violates energy conservation since higher potential differences always accelerate electrons to higher speeds. Remember this direct proportionality: doubling voltage doubles the kinetic energy gained. This relationship holds for any charged particle in non-relativistic motion, making it a fundamental pattern in electrostatics problems.
A point charge $+Q$ is fixed at the origin. A test charge $+q$ is moved from point A (at distance $r$ from the origin) to point B (at distance $3r$ from the origin) along a straight radial path. How much work is done by an external agent to move the charge at constant speed?
Answer: $k\frac{Qq}{3r} - k\frac{Qq}{r} = k\frac{Qq}{r}\left(\frac{1}{3} - 1\right)$ When you encounter problems involving moving charges in electric fields, think about electric potential energy and the work-energy theorem. The key insight is that work done by an external agent equals the change in potential energy when moving at constant speed. The electric potential energy of a test charge $+q$ at distance $r$ from a point charge $+Q$ is $U = k\frac{Qq}{r}$ . Since both charges are positive, this energy is positive and decreases as distance increases. Initially at point A (distance $r$ ): $U_A = k\frac{Qq}{r}$ Finally at point B (distance $3r$ ): $U_B = k\frac{Qq}{3r}$ The work done by the external agent equals the change in potential energy: $W_{ext} = U_B - U_A = k\frac{Qq}{3r} - k\frac{Qq}{r}$ . This gives us $k\frac{Qq}{r}\left(\frac{1}{3} - 1\right)$ , which matches choice B. Choice A incorrectly calculates $U_A - U_B$ instead of $U_B - U_A$ . This would be the work done by the electric field, not the external agent. Choice C adds the energies instead of finding their difference, which doesn't represent any meaningful physical quantity. Choice D uses $9r$ instead of $3r$ in the denominator, possibly confusing the distance with distance squared. Remember: when moving charges at constant speed, the external agent does work equal to the change in potential energy. Always subtract initial from final energy, and be careful about which agent (external or field) the problem asks about.
A dipole consisting of charges $+q$ and $-q$ separated by distance $d$ is initially oriented perpendicular to a uniform electric field $E$ . The dipole is then rotated to align with the field. If the external work done during this rotation is $W_{ext} = +qEd$ , what is the change in the system's electric potential energy?
Answer: $-qEd$ because the dipole moves to a lower potential energy configuration When analyzing dipole behavior in electric fields, you need to consider both the work-energy theorem and the direction of spontaneous motion. A dipole naturally wants to align with an external electric field because this represents the lowest potential energy state. Since the dipole starts perpendicular to the field and rotates to align with it, this is energetically favorable - the system moves from higher to lower potential energy. The potential energy change is $\Delta U = -qEd$ , meaning the potential energy decreases by $qEd$ . To verify this using the work-energy theorem: $W_{ext} + W_{field} = \Delta U$ . The external work is $+qEd$ , and since the field does negative work when external forces rotate the dipole against the field's natural tendency, we get $qEd + (-qEd) = 0$ . Wait - this seems wrong. Actually, when rotating to the favorable position, the field does positive work, so $W_{field} = +qEd$ . This gives us $\Delta U = W_{ext} + W_{field} = qEd + (-2qEd) = -qEd$ . Answer A incorrectly assumes external work always increases potential energy, but the relationship depends on whether other forces (like the field) also do work. Answer B wrongly suggests that distance from the field source matters, but uniform fields have constant strength everywhere. Answer C miscalculates by double-counting the charge contribution. Remember: when a dipole aligns with a field naturally, potential energy decreases. The work-energy theorem accounts for ALL forces doing work, not just external ones.
Two parallel plates separated by distance $d$ have a potential difference of $V_0$ . A proton (charge $+e$ , mass $m$ ) is released from rest at the positive plate. When it reaches the negative plate, what fraction of its kinetic energy would be converted to electric potential energy if it were instead stopped at the midpoint between the plates?
Answer: $\frac{1}{2}$ because the proton travels half the distance and potential varies linearly When the proton reaches the negative plate, its kinetic energy is $K = eV_0$ . At the midpoint, the potential is $\frac{V_0}{2}$ (linear variation between plates), so if stopped there, the potential energy would be $U = e \cdot \frac{V_0}{2} = \frac{eV_0}{2}$ . The fraction is $\frac{U}{K} = \frac{eV_0/2}{eV_0} = \frac{1}{2}$ . Choice A incorrectly assumes quadratic potential variation. Choice C confuses momentum with energy conservation. Choice D misunderstands that we're comparing energies at different positions, not asking about total energy conversion.
A charge of $+3.0 \times 10^{-6}$ C is moved from point A to point B in a uniform electric field. The electric potential at point A is $+50$ V and at point B is $+20$ V. What is the change in electric potential energy of the charge?
Answer: $-9.0 \times 10^{-5}$ J When you encounter problems involving electric potential and potential energy, remember that these are closely related but distinct concepts. Electric potential (measured in volts) is a property of the electric field at a point, while electric potential energy depends on both the field and the specific charge present. The relationship between change in potential energy and electric potential is: $\Delta U = q \Delta V$ , where $q$ is the charge and $\Delta V$ is the change in electric potential. Here, the change in potential is $\Delta V = V_B - V_A = 20\text{ V} - 50\text{ V} = -30\text{ V}$ . Therefore: $\Delta U = (3.0 \times 10^{-6}\text{ C})(-30\text{ V}) = -9.0 \times 10^{-5}\text{ J}$ The negative sign indicates that the potential energy decreases as the positive charge moves from higher potential to lower potential, which makes physical sense—positive charges naturally move toward lower potential regions when free to do so. Answer A gives the correct magnitude but wrong sign—this would occur if you calculated $V_A - V_B$ instead of $V_B - V_A$ . Answers C and D both give $\pm 2.1 \times 10^{-4}\text{ J}$ , which you'd get by incorrectly adding the potentials ( $50 + 20 = 70$ ) instead of finding their difference. Answer D has the right sign but wrong magnitude from this addition error. Remember: potential energy change equals charge times potential difference, and always subtract initial from final values ( $\Delta V = V_{final} - V_{initial}$ ) to get the correct sign. The sign tells you whether energy increases or decreases during the process.
A positive charge $+q$ is moved from point A to point B in an electric field, where the electric potential at A is $V_A = 20\text{ V}$ and at B is $V_B = 8\text{ V}$ . If the charge then moves from B to point C where $V_C = 15\text{ V}$ , what is the total change in electric potential energy for the complete journey from A to C?
Answer: $-5q\text{ J}$ because the net potential difference is $15\text{ V} - 20\text{ V} = -5\text{ V}$ The change in electric potential energy is $\Delta U = q(V_C - V_A) = q(15\text{ V} - 20\text{ V}) = -5q\text{ J}$ . Electric potential energy depends only on the initial and final positions, not on the path taken. The intermediate stop at point B is irrelevant for calculating the total energy change. Choice B incorrectly assumes the charge gains energy when moving to higher potential (it actually loses energy when moving from high to low potential). Choice C incorrectly adds magnitudes rather than using the net potential difference. Choice D incorrectly assumes that being at an intermediate potential means no energy change occurred.
An alpha particle (charge $+2e$ ) is fired directly toward a stationary gold nucleus (charge $+79e$ ) from a great distance with initial kinetic energy $K_0$ . At the point of closest approach, what fraction of the initial kinetic energy has been converted to electric potential energy?
Answer: $1$ When an alpha particle approaches a gold nucleus, you're dealing with a classic Coulomb scattering problem that tests conservation of energy. At the point of closest approach, the alpha particle momentarily stops before being repelled back, meaning all its kinetic energy has been converted to electric potential energy. Using conservation of energy, the initial kinetic energy $K_0$ equals the sum of kinetic and potential energies at any point: $K_0 = K + U$ . At closest approach, the velocity becomes zero, so $K = 0$ and therefore $K_0 = U$ . This means the electric potential energy equals the initial kinetic energy, making the fraction $\frac{U}{K_0} = \frac{K_0}{K_0} = 1$ . Answer D (1) is correct because 100% of the initial kinetic energy converts to potential energy at closest approach. Answer A ( $\frac{1}{4}$ ) might come from incorrectly applying inverse-square relationships or confusing this with orbital mechanics problems. Answer B ( $\frac{1}{2}$ ) could result from incorrectly assuming the particle retains half its kinetic energy at closest approach, perhaps by analogy with other collision problems. Answer C ( $\frac{3}{4}$ ) might stem from partial application of energy conservation or confusion about when maximum potential energy occurs. Remember: at the turning point in any conservative force problem, kinetic energy reaches zero and potential energy reaches its maximum. For Coulomb scattering, this maximum potential energy always equals the initial kinetic energy, regardless of the specific charges involved.
A proton starts from rest and accelerates through a potential difference of $1000$ V. Using conservation of energy, what is the final speed of the proton? (Mass of proton = $1.67 \times 10^{-27}$ kg, charge = $+1.6 \times 10^{-19}$ C)
Answer: $4.4 \times 10^5$ m/s When you see a charged particle accelerating through a potential difference, you're dealing with energy conservation. The electrical potential energy lost by the particle converts directly into kinetic energy. Set up the energy conservation equation: the initial energy equals the final energy. Since the proton starts from rest, its initial kinetic energy is zero, so all the electrical potential energy becomes kinetic energy: $qV = \frac{1}{2}mv^2$ Solving for velocity: $v = \sqrt{\frac{2qV}{m}}$ Substitute the given values: $v = \sqrt{\frac{2(1.6 \times 10^{-19})(1000)}{1.67 \times 10^{-27}}}$ $v = \sqrt{\frac{3.2 \times 10^{-16}}{1.67 \times 10^{-27}}} = \sqrt{1.92 \times 10^{11}} = 4.4 \times 10^5 \text{ m/s}$ This confirms answer A is correct. Answer B ( $2.2 \times 10^5$ m/s) results from forgetting the factor of 2 in the kinetic energy formula, using $qV = mv^2$ instead of $qV = \frac{1}{2}mv^2$ . Answer C ( $6.2 \times 10^5$ m/s) likely comes from calculation errors or using incorrect values. Answer D ( $3.1 \times 10^5$ m/s) could result from mixing up the mass or charge values during calculation. Remember: for any charged particle problem involving potential differences, energy conservation with $qV = \frac{1}{2}mv^2$ is your go-to approach. Always double-check that factor of $\frac{1}{2}$ in kinetic energy—it's a common source of errors.
A capacitor with capacitance $C$ is charged to voltage $V_0$ . The energy stored is initially $U_0 = \frac{1}{2}CV_0^2$ . The capacitor is then connected to an identical uncharged capacitor. After equilibrium is reached, what is the total energy stored in both capacitors?
Answer: $\frac{U_0}{2}$ When two capacitors are connected, you're dealing with both charge conservation and energy considerations. This type of problem tests whether you understand that energy can be "lost" during charge redistribution, even though charge is conserved. Initially, the charged capacitor has charge $Q_0 = CV_0$ and energy $U_0 = \frac{1}{2}CV_0^2$ . When connected to an identical uncharged capacitor, charge flows until both capacitors reach the same voltage. Since charge is conserved, the total charge $Q_0$ gets distributed equally between the two identical capacitors: each gets $\frac{Q_0}{2}$ . The final voltage across each capacitor is $V_f = \frac{Q_0/2}{C} = \frac{V_0}{2}$ . The energy stored in each capacitor is now $\frac{1}{2}C(\frac{V_0}{2})^2 = \frac{1}{8}CV_0^2 = \frac{U_0}{4}$ . The total energy in both capacitors is $2 \times \frac{U_0}{4} = \frac{U_0}{2}$ , confirming answer B. Answer A ( $\frac{U_0}{4}$ ) gives only the energy in one capacitor, not both. Answer C ( $U_0$ ) incorrectly assumes energy is conserved - this would violate the second law of thermodynamics since energy is dissipated as heat during the charge redistribution. Answer D ( $2U_0$ ) impossibly suggests energy creation from nothing. Remember: in capacitor problems involving charge redistribution, charge is always conserved but energy typically decreases due to resistive losses in the connecting wires. Always check both conservation laws separately.
A charged particle oscillates in a one-dimensional electric potential well described by $V(x) = \frac{1}{2}kx^2$ where $k > 0$ . If the particle has charge $q > 0$ and total mechanical energy $E$ , what is the amplitude of oscillation?
Answer: $A = \sqrt{\frac{2E}{qk}}$ When you encounter a charged particle in a potential well, you're dealing with conservation of energy and the relationship between electric potential energy and force. The key insight is understanding how the given potential relates to the actual force on the charged particle. The potential energy of the charged particle is $U(x) = qV(x) = q \cdot \frac{1}{2}kx^2 = \frac{1}{2}qkx^2$ . This gives us a simple harmonic oscillator with an effective spring constant of $qk$ . At the turning points (maximum displacement ±A), all energy is potential energy since the particle momentarily stops. Using conservation of energy: $E = U_{max} = \frac{1}{2}qkA^2$ . Solving for amplitude: $A = \sqrt{\frac{2E}{qk}}$ , which is answer D. The wrong answers represent common mistakes: Answer A ( $\sqrt{\frac{E}{k}}$ ) incorrectly uses the given potential $V(x)$ directly instead of the actual potential energy $U(x) = qV(x)$ , and misses the factor of 2 from the energy equation. Answer B ( $\sqrt{\frac{2E}{k}}$ ) correctly includes the factor of 2 but still ignores the charge $q$ . Answer C ( $\sqrt{\frac{E}{qk}}$ ) properly includes the charge but incorrectly omits the factor of 2 that comes from $\frac{1}{2}qkA^2 = E$ . Remember: always distinguish between electric potential $V(x)$ and potential energy $U(x) = qV(x)$ . The charge multiplies the given potential to get the actual energy stored in the system.

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