1. In electrostatics, a charge is moved from point \(P\) to point \(Q\) along three different paths in the field of fixed source charges. Which statement best expresses the conservative nature of the electrostatic force?
ⓐ. The work done by the electrostatic force is always zero for any displacement
ⓑ. The work done by the electrostatic force depends only on the speed of the moving charge
ⓒ. Work depends only on points \(P\) and \(Q\), not on the path
ⓓ. The work done by the electrostatic force is maximum along the shortest path
Correct Answer: Work depends only on points \(P\) and \(Q\), not on the path
Explanation: A conservative force is identified by path independence of work. In an electrostatic field produced by fixed charges, the work done in moving a test charge from \(P\) to \(Q\) is determined by the initial and final positions. The actual curved, straight, long, or short route between the two points does not change the work done by the electrostatic force. This property allows electrostatic potential energy to be defined consistently because the energy change is not route-dependent. The work is not always zero; it is zero only for a closed path or for movement between points of equal potential in later terminology. A common mistake is to connect “shortest path” with “least work,” but that idea does not apply to a conservative electrostatic force.
2. Assertion: The work done by electrostatic force in moving a charge around any closed path is zero.
Reason: Electrostatic force is conservative, so the work done depends only on the initial and final positions.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In a closed path, the initial and final positions of the charge are the same. For a conservative force, work depends only on these two endpoint positions, not on the route taken. Since the endpoint is unchanged in a closed loop, the change in electrostatic potential energy is zero. Using the relation \(\Delta U=-W_{\text{field}}\), if \(\Delta U=0\), then \(W_{\text{field}}=0\). Therefore, the Reason gives the exact logic behind the Assertion. The important condition is that the electrostatic field is produced by stationary source charges; time-varying electromagnetic situations are not being considered here.
3. A charge is moved from point \(A\) to point \(B\) in an electrostatic field. The electrostatic force does \(+8.0\,\text{mJ}\) of work along Path P. What is the work done by the electrostatic force if the same charge is moved from \(A\) to \(B\) along Path Q, and then from \(B\) back to \(A\) along any path?
ⓐ. Along Path Q: \(+8.0\,\text{mJ}\); from \(B\) to \(A\): \(-8.0\,\text{mJ}\)
ⓑ. Along Path Q: \(-8.0\,\text{mJ}\); from \(B\) to \(A\): \(+8.0\,\text{mJ}\)
ⓒ. Along Path Q: \(+16.0\,\text{mJ}\); from \(B\) to \(A\): \(0\,\text{mJ}\)
ⓓ. Along Path Q: \(0\,\text{mJ}\); from \(B\) to \(A\): \(+8.0\,\text{mJ}\)
Correct Answer: Along Path Q: \(+8.0\,\text{mJ}\); from \(B\) to \(A\): \(-8.0\,\text{mJ}\)
Explanation: \( \textbf{Given:} \) Work done by the electrostatic force from \(A\) to \(B\) along Path P is \(W_{A\to B}=+8.0\,\text{mJ}\).
\( \textbf{Key property:} \) Electrostatic force is conservative, so \(W_{A\to B}\) is path independent.
\( \textbf{Same endpoints:} \) Path Q also starts at \(A\) and ends at \(B\), so the work remains \(+8.0\,\text{mJ}\).
\( \textbf{Reverse motion:} \) Moving from \(B\) to \(A\) reverses the endpoints.
\( \textbf{Work reversal:} \) For a conservative force, \(W_{B\to A}=-W_{A\to B}\).
\( \textbf{Substitution:} \) \(W_{B\to A}=-(+8.0\,\text{mJ})=-8.0\,\text{mJ}\).
\( \textbf{Closed-loop check:} \) \(W_{A\to B}+W_{B\to A}=+8.0\,\text{mJ}-8.0\,\text{mJ}=0\).
\( \textbf{Final answer:} \) Along Path Q the work is \(+8.0\,\text{mJ}\), and from \(B\) to \(A\) it is \(-8.0\,\text{mJ}\).
4. Complete the statement correctly: The existence of a well-defined electrostatic potential energy is possible because the work done by electrostatic force is ______.
ⓐ. always positive and proportional to time taken
ⓑ. path independent and zero over a closed path
ⓒ. dependent on path and independent of endpoints
ⓓ. independent of charge and dependent only on mass
Correct Answer: path independent and zero over a closed path
Explanation: Electrostatic potential energy can be assigned to a charge configuration only when the work-energy change between two states is definite. This is possible because electrostatic force is conservative. A conservative force does path-independent work between two points. For a closed path, the initial and final positions are the same, so the net work done by the electrostatic force is zero. If work depended on the route, the same final position could have different potential energy values, which would make potential energy ill-defined. The important distinction is that “conservative” does not mean “work is always zero”; it means the work is route-independent.
5. Study the table and identify the row that correctly compares a conservative electrostatic force with a non-conservative force.
| Row | Electrostatic conservative force | Non-conservative force |
|---|
| P | Work depends on path length | Work depends only on endpoints |
| Q | Closed-loop work is zero | Closed-loop work can be non-zero |
| R | No potential energy can be defined | Potential energy is always uniquely defined |
| S | Work is always negative | Work is always positive |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row S
ⓓ. Row R
Correct Answer: Row Q
Explanation: For an electrostatic force produced by fixed charges, the field is conservative. A conservative force has zero work over any closed path because the initial and final positions coincide. Non-conservative forces, such as friction, can do non-zero work over a closed path because their work may depend on the actual path length or route. Row P reverses the correct comparison. Row R is wrong because potential energy can be defined for conservative electrostatic force. Row S is also wrong because the work done by electrostatic force may be positive, negative, or zero depending on the displacement and charge situation. The reliable test is closed-loop work, not the sign of work in one particular motion.
6. Use the situation below and answer the question.
A charge is taken from point \(M\) to point \(N\) in the field of fixed source charges. Along Route I, the electrostatic force does \(+5.0\,\text{J}\) of work. Along Route II, an extra curved detour is added, but the starting and ending points remain \(M\) and \(N\).
Which conclusion is correct for the work done by the electrostatic force along Route II?
ⓐ. It becomes \(0\,\text{J}\) because the path is curved
ⓑ. It is less than \(+5.0\,\text{J}\) because Route II is longer
ⓒ. It is greater than \(+5.0\,\text{J}\) because Route II has an extra detour
ⓓ. It is exactly \(+5.0\,\text{J}\) because the endpoints are unchanged
Correct Answer: It is exactly \(+5.0\,\text{J}\) because the endpoints are unchanged
Explanation: The field of fixed source charges is an electrostatic field, so the electrostatic force is conservative. For a conservative force, the work done between two points is independent of the path followed. Route II may be longer and more curved than Route I, but it still begins at \(M\) and ends at \(N\). Therefore, the work done by the electrostatic force remains \(+5.0\,\text{J}\). The extra detour would matter for a path-dependent force such as friction, but not for an electrostatic conservative force. A curved path does not automatically imply zero work; zero work is guaranteed for a complete closed path in electrostatics.
7. Consider the following statements about work done by electrostatic force.
Statement I: If a charge returns to its starting point, the total electrostatic work done over the complete path is zero.
Statement II: If two paths have the same initial and final points, electrostatic work is the same along both paths.
Statement III: Electrostatic work depends on the time taken to move the charge slowly between two fixed points.
Which statements are correct?
ⓐ. I and III only
ⓑ. I, II and III
ⓒ. I and II only
ⓓ. II and III only
Correct Answer: I and II only
Explanation: Statement I is correct because a closed path has the same initial and final position. Since electrostatic force is conservative, the net work over such a closed path is zero. Statement II is also correct because path independence is the defining work property of a conservative force. Statement III is incorrect in the electrostatic work idea used here because the work done by the electrostatic force between two fixed points is not decided by the time taken. The time taken may affect power or the details of an external mechanism, but it does not change the electrostatic work between fixed endpoints. The usual confusion is to mix “slow movement” used in later external-work definitions with the conservative-force property itself.
8. A charge is moved in an electrostatic field from point \(X\) to point \(Y\). The change in its electrostatic potential energy is \(+12\,\text{J}\). What is the work done by the electrostatic force during this movement, and what is the work over the return path from \(Y\) to \(X\)?
ⓐ. \(W_{X\to Y}=+12\,\text{J}\) and \(W_{Y\to X}=-12\,\text{J}\)
ⓑ. \(W_{X\to Y}=0\,\text{J}\) and \(W_{Y\to X}=+12\,\text{J}\)
ⓒ. \(W_{X\to Y}=+24\,\text{J}\) and \(W_{Y\to X}=0\,\text{J}\)
ⓓ. \(W_{X\to Y}=-12\,\text{J}\) and \(W_{Y\to X}=+12\,\text{J}\)
Correct Answer: \(W_{X\to Y}=-12\,\text{J}\) and \(W_{Y\to X}=+12\,\text{J}\)
Explanation: \( \textbf{Given:} \) Change in electrostatic potential energy for motion from \(X\) to \(Y\) is \(\Delta U=+12\,\text{J}\).
\( \textbf{Required:} \) Work done by the electrostatic force for \(X\to Y\) and for the reverse motion \(Y\to X\).
\( \textbf{Work-energy relation:} \)
\[
\Delta U=-W_{\text{field}}
\]
\( \textbf{Why this applies:} \) Electrostatic force is conservative, so its work is linked to change in potential energy.
\( \textbf{For motion from \(X\) to \(Y\):} \)
\[
W_{\text{field}}=-\Delta U=-(+12\,\text{J})=-12\,\text{J}
\]
\( \textbf{Reverse path logic:} \) In moving from \(Y\) to \(X\), the potential energy change becomes \(-12\,\text{J}\).
\( \textbf{Reverse work:} \)
\[
W_{Y\to X}=-(-12\,\text{J})=+12\,\text{J}
\]
\( \textbf{Closed-loop check:} \) \((-12\,\text{J})+(+12\,\text{J})=0\,\text{J}\), matching the conservative nature of electrostatic force.
\( \textbf{Final answer:} \) \(W_{X\to Y}=-12\,\text{J}\) and \(W_{Y\to X}=+12\,\text{J}\).
9. Match the ideas in Column I with their correct meanings in Column II.
| Column I | Column II |
|---|
| P. Conservative electrostatic force | 1. Work may depend on the actual route followed |
| Q. Closed path in electrostatics | 2. Work depends only on initial and final positions |
| R. Non-conservative force | 3. Net electrostatic work is zero |
| S. Potential energy relation | 4. \(\Delta U=-W_{\text{field}}\) |
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-2, Q-1, R-3, S-4
ⓒ. P-4, Q-3, R-2, S-1
ⓓ. P-3, Q-2, R-1, S-4
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: A conservative electrostatic force is one for which work depends only on the initial and final positions, so P matches 2. For a closed path, the starting and ending positions are the same, so the net electrostatic work is zero; Q matches 3. A non-conservative force can have path-dependent work, so R matches 1. The change in electrostatic potential energy is related to field work by \(\Delta U=-W_{\text{field}}\), so S matches 4. This matching also shows why potential energy is introduced only after establishing conservative work. The sign in \(\Delta U=-W_{\text{field}}\) is an important point: positive work by the field lowers potential energy.
10. A graph is described for a charge moving from point \(A\) to point \(B\) in an electrostatic field. The horizontal axis represents different paths \(P\), \(Q\), and \(R\), all having the same endpoints \(A\) and \(B\). The vertical axis represents work \(W_{\text{field}}\) done by the electrostatic force. Which graph behaviour is physically correct?
ⓐ. The bar for the longest path is highest
ⓑ. The bar for the shortest path is highest
ⓒ. Three bars of equal height for \(P\), \(Q\), and \(R\)
ⓓ. All bars are zero for any two different endpoints
Correct Answer: Three bars of equal height for \(P\), \(Q\), and \(R\)
Explanation: \( \textbf{Graph interpretation:} \) The graph compares \(W_{\text{field}}\) for different paths with the same endpoints \(A\) and \(B\).
\( \textbf{Conservative-force rule:} \) In electrostatics, \(W_{\text{field}}\) is independent of path.
\( \textbf{Endpoint condition:} \) Since paths \(P\), \(Q\), and \(R\) all start at \(A\) and end at \(B\), the electrostatic work must be identical for all three.
\( \textbf{Correct graph shape:} \) Equal work values mean equal-height bars.
\( \textbf{Why shortest path is not special:} \) Path length does not decide electrostatic work between fixed endpoints.
\( \textbf{Why zero is not guaranteed:} \) Work is guaranteed to be zero for a closed path, not for every pair of different endpoints.
\( \textbf{Physical meaning:} \) The graph would visually demonstrate path independence, which is the operational test of a conservative electrostatic force.
\( \textbf{Final answer:} \) The correct graph has three bars of equal height for \(P\), \(Q\), and \(R\).
11. A positive charge is moved between two points in an electrostatic field, and the work done by the electrostatic force is negative. What does this imply about the electrostatic potential energy of the charge?
ⓐ. The electrostatic potential energy increases
ⓑ. The electrostatic potential energy decreases
ⓒ. The electrostatic potential energy becomes zero at the final point
ⓓ. The electrostatic potential energy remains unchanged
Correct Answer: The electrostatic potential energy increases
Explanation: The change in electrostatic potential energy is related to the work done by the electrostatic field by \(\Delta U=-W_{\text{field}}\). If \(W_{\text{field}}\) is negative, then \(\Delta U\) becomes positive. A positive \(\Delta U\) means the final electrostatic potential energy is greater than the initial electrostatic potential energy. Physically, negative work by the field means the motion is against the natural electrostatic tendency of the charge. In such a case, an external agent must usually supply energy if the charge is moved slowly. The sign of work, not the sign of the charge alone, decides whether \(U\) increases or decreases.
12. A charge is moved slowly from point \(P\) to point \(Q\) in an electrostatic field without any change in kinetic energy. The electrostatic field does \(-6.0\,\text{J}\) of work. What are the change in electrostatic potential energy and the work done by the external agent?
ⓐ. \(\Delta U=0\,\text{J}\), \(W_{\text{ext}}=+6.0\,\text{J}\)
ⓑ. \(\Delta U=+6.0\,\text{J}\), \(W_{\text{ext}}=+6.0\,\text{J}\)
ⓒ. \(\Delta U=-6.0\,\text{J}\), \(W_{\text{ext}}=-6.0\,\text{J}\)
ⓓ. \(\Delta U=+6.0\,\text{J}\), \(W_{\text{ext}}=-6.0\,\text{J}\)
Correct Answer: \(\Delta U=+6.0\,\text{J}\), \(W_{\text{ext}}=+6.0\,\text{J}\)
Explanation: \( \textbf{Given:} \) Work done by the electrostatic field is \(W_{\text{field}}=-6.0\,\text{J}\).
\( \textbf{Condition:} \) The charge is moved slowly, so the change in kinetic energy is zero.
\( \textbf{Potential-energy relation:} \)
\[
\Delta U=-W_{\text{field}}
\]
\( \textbf{Substitution:} \)
\[
\Delta U=-(-6.0\,\text{J})=+6.0\,\text{J}
\]
\( \textbf{External-work idea:} \) For slow movement, the external agent balances the electric force so that no kinetic energy is gained.
\( \textbf{Work by external agent:} \)
\[
W_{\text{ext}}=\Delta U
\]
\( \textbf{Final calculation:} \)
\[
W_{\text{ext}}=+6.0\,\text{J}
\]
\( \textbf{Sign check:} \) The field opposes the motion, so the external agent must do positive work.
\( \textbf{Final answer:} \) \(\Delta U=+6.0\,\text{J}\) and \(W_{\text{ext}}=+6.0\,\text{J}\).
13. Assertion: If the electrostatic field does positive work on a charge, the electrostatic potential energy of the charge decreases.
Reason: In electrostatics, \(\Delta U=-W_{\text{field}}\).
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Electrostatic potential energy is connected with the work done by the electrostatic field through \(\Delta U=-W_{\text{field}}\). If \(W_{\text{field}}\) is positive, then \(\Delta U\) is negative. A negative \(\Delta U\) means the final potential energy is less than the initial potential energy. This is the same energy logic seen in many conservative-force situations: when the field itself helps the motion, stored potential energy is converted into some other form such as kinetic energy. The Reason is not merely a related formula; it directly proves the Assertion. A common sign-related mistake is to assume positive field work always means positive energy change.
14. A small charge is released from rest at point \(X\) in an electrostatic field and later passes point \(Y\). Neglect non-electrostatic forces. Which statement correctly compares the work done by the field and the change in electrostatic potential energy?
ⓐ. \(W_{\text{field}}\) is negative and \(\Delta U\) is positive
ⓑ. \(W_{\text{field}}\) is positive and \(\Delta U\) is positive
ⓒ. \(W_{\text{field}}\) is positive and \(\Delta U\) is negative
ⓓ. \(W_{\text{field}}\) is zero and \(\Delta U\) is negative
Correct Answer: \(W_{\text{field}}\) is positive and \(\Delta U\) is negative
Explanation: When the charge is released from rest and moves only due to electrostatic force, the electrostatic field does work on it in the direction of its motion. The charge gains kinetic energy, so the work done by the field is positive. For a conservative electrostatic field, this positive field work comes from a decrease in electrostatic potential energy. The relation \(\Delta U=-W_{\text{field}}\) shows that if \(W_{\text{field}}\gt0\), then \(\Delta U\lt0\). This does not require knowing whether the charge is positive or negative; the actual motion already tells us the field is doing positive work. A common mistake is to connect the sign of \(\Delta U\) directly to the sign of charge instead of the sign of field work.
15. A charge is moved slowly in an electrostatic field through a process described in the table.
| Case | Work done by electrostatic field | Movement condition |
|---|
| P | \(+4\,\text{J}\) | Slow movement |
| Q | \(-4\,\text{J}\) | Slow movement |
| R | \(0\,\text{J}\) | Slow movement |
Which row correctly gives the work done by the external agent?
ⓐ. \(P:-4\,\text{J}\), \(Q:+4\,\text{J}\), \(R:0\,\text{J}\)
ⓑ. \(P:+8\,\text{J}\), \(Q:-8\,\text{J}\), \(R:0\,\text{J}\)
ⓒ. \(P:0\,\text{J}\), \(Q:+4\,\text{J}\), \(R:-4\,\text{J}\)
ⓓ. \(P:+4\,\text{J}\), \(Q:-4\,\text{J}\), \(R:0\,\text{J}\)
Correct Answer: \(P:-4\,\text{J}\), \(Q:+4\,\text{J}\), \(R:0\,\text{J}\)
Explanation: \( \textbf{Slow-movement condition:} \) If the charge is moved slowly, its kinetic energy does not change.
\( \textbf{Work balance:} \) With no change in kinetic energy, the net work is zero, so \(W_{\text{ext}}+W_{\text{field}}=0\).
\( \textbf{External-work relation:} \)
\[
W_{\text{ext}}=-W_{\text{field}}
\]
\( \textbf{Case P:} \) If \(W_{\text{field}}=+4\,\text{J}\), then \(W_{\text{ext}}=-4\,\text{J}\).
\( \textbf{Case Q:} \) If \(W_{\text{field}}=-4\,\text{J}\), then \(W_{\text{ext}}=+4\,\text{J}\).
\( \textbf{Case R:} \) If \(W_{\text{field}}=0\,\text{J}\), then \(W_{\text{ext}}=0\,\text{J}\).
\( \textbf{Energy check:} \) The same result follows from \(\Delta U=-W_{\text{field}}\) and \(W_{\text{ext}}=\Delta U\) for slow movement.
\( \textbf{Final answer:} \) \(P:-4\,\text{J}\), \(Q:+4\,\text{J}\), and \(R:0\,\text{J}\).
16. For a slowly moved charge in an electrostatic field, the blank in the statement should be filled correctly: If the external agent does positive work, the electrostatic potential energy of the charge ______.
ⓐ. decreases because the field must have done positive work
ⓑ. increases by the external work done in slow motion
ⓒ. becomes independent of the initial and final positions
ⓓ. remains unchanged because slow motion means no work is done
Correct Answer: increases by the external work done in slow motion
Explanation: Slow movement means the charge is not allowed to gain kinetic energy. Under this condition, the work done by the external agent changes the electrostatic potential energy. If \(W_{\text{ext}}\) is positive, then \(\Delta U\) is positive for slow movement. This means the charge is being taken to a configuration of higher electrostatic potential energy. The electrostatic field would do negative work during such a process because \(W_{\text{field}}=-\Delta U\). The word “slow” does not mean no work; it means no net kinetic-energy change.
17. Read the situation below and answer the question.
A positive charge \(+q\) is held near a fixed positive source charge. It is then moved slowly farther away from the source charge by an external agent.
Which statement best describes the work and electrostatic potential energy during this movement?
ⓐ. The electrostatic field does positive work and \(U\) increases
ⓑ. The electrostatic field does negative work and \(U\) decreases
ⓒ. The external agent does negative work and \(U\) decreases
ⓓ. The external agent does positive work and \(U\) increases
Correct Answer: The external agent does negative work and \(U\) decreases
Explanation: Two positive charges repel each other, so the electrostatic force on \(+q\) is away from the fixed positive source charge. When \(+q\) is moved farther away, its displacement is along the electrostatic force. Therefore, the electrostatic field does positive work. From \(\Delta U=-W_{\text{field}}\), positive field work means the electrostatic potential energy decreases. For slow movement, the external agent must oppose the natural acceleration and do negative work. This situation is often misunderstood because “external agent moves the charge” does not automatically mean the external agent does positive work; the sign depends on the direction of its force relative to displacement.
18. A negative charge is placed near a fixed positive source charge and is moved slowly farther away from it by an external agent. Which statement is correct?
ⓐ. The electrostatic field does positive work and \(U\) decreases
ⓑ. The external agent does positive work and \(U\) increases
ⓒ. The external agent does negative work and \(U\) decreases
ⓓ. The electrostatic field does negative work and \(U\) increases
Correct Answer: The external agent does positive work and \(U\) increases
Explanation: A negative charge is attracted toward a positive source charge. If it is moved farther away, its displacement is opposite to the electrostatic force acting on it. Therefore, the electrostatic field does negative work. Using \(\Delta U=-W_{\text{field}}\), a negative value of \(W_{\text{field}}\) gives a positive \(\Delta U\). The electrostatic potential energy increases because the charge is being separated against attraction. Since the movement is slow, the external agent supplies the energy, so \(W_{\text{ext}}=\Delta U\gt0\). A common misconception is that a negative charge must always have decreasing potential energy when it moves, but the direction of movement relative to force decides the energy change.
19. A charge is moved slowly between two points in an electrostatic field. Which pair of relations is correct for this slow movement?
ⓐ. \(\Delta U=-W_{\text{ext}}\) and \(W_{\text{field}}=\Delta U\)
ⓑ. \(\Delta U=-W_{\text{field}}\) and \(W_{\text{ext}}=\Delta U\)
ⓒ. \(\Delta U=W_{\text{ext}}\) and \(W_{\text{field}}=W_{\text{ext}}\)
ⓓ. \(\Delta U=W_{\text{field}}\) and \(W_{\text{ext}}=-\Delta U\)
Correct Answer: \(\Delta U=-W_{\text{field}}\) and \(W_{\text{ext}}=\Delta U\)
Explanation: The relation \(\Delta U=-W_{\text{field}}\) is the basic work-energy relation for a conservative electrostatic force. It says that if the field does positive work, potential energy decreases, and if the field does negative work, potential energy increases. For slow movement, the charge has no change in kinetic energy. Hence the external work must account for the change in electrostatic potential energy, giving \(W_{\text{ext}}=\Delta U\). This also implies \(W_{\text{ext}}=-W_{\text{field}}\) during slow movement. The most common sign error is to write \(\Delta U=W_{\text{field}}\), which reverses the energy conversion.
20. A charge is moved slowly from position \(R\) to position \(S\). The external agent does \(+15\,\text{mJ}\) of work. Which statement correctly gives the work done by the electrostatic field and the change in electrostatic potential energy?
ⓐ. \(W_{\text{field}}=0\,\text{mJ}\), \(\Delta U=+15\,\text{mJ}\)
ⓑ. \(W_{\text{field}}=-30\,\text{mJ}\), \(\Delta U=+15\,\text{mJ}\)
ⓒ. \(W_{\text{field}}=-15\,\text{mJ}\), \(\Delta U=+15\,\text{mJ}\)
ⓓ. \(W_{\text{field}}=+15\,\text{mJ}\), \(\Delta U=-15\,\text{mJ}\)
Correct Answer: \(W_{\text{field}}=-15\,\text{mJ}\), \(\Delta U=+15\,\text{mJ}\)
Explanation: \( \textbf{Given:} \) Work done by external agent is \(W_{\text{ext}}=+15\,\text{mJ}\).
\( \textbf{Movement condition:} \) The movement is slow, so \(\Delta K=0\).
\( \textbf{Work balance:} \)
\[
W_{\text{ext}}+W_{\text{field}}=\Delta K=0
\]
\( \textbf{Field work:} \)
\[
W_{\text{field}}=-W_{\text{ext}}=-15\,\text{mJ}
\]
\( \textbf{Potential-energy relation:} \)
\[
\Delta U=-W_{\text{field}}
\]
\( \textbf{Substitution:} \)
\[
\Delta U=-(-15\,\text{mJ})=+15\,\text{mJ}
\]
\( \textbf{Physical meaning:} \) Positive external work is stored as increased electrostatic potential energy.
\( \textbf{Unit check:} \) Work and potential energy are both measured in \(\text{J}\) or \(\text{mJ}\).
\( \textbf{Final answer:} \) \(W_{\text{field}}=-15\,\text{mJ}\) and \(\Delta U=+15\,\text{mJ}\).
