101. In a uniform electric field \(\vec{E}\), a displacement \(\vec{d}\) is made exactly opposite to the field direction. Which expression gives \(V_B-V_A\) for displacement from \(A\) to \(B\)?
ⓐ. \(V_B-V_A=-Ed\)
ⓑ. \(V_B-V_A=-\frac{E}{d}\)
ⓒ. \(V_B-V_A=0\)
ⓓ. \(V_B-V_A=+Ed\)
Correct Answer: \(V_B-V_A=+Ed\)
Explanation: The general relation for a uniform electric field is \(V_B-V_A=-\vec{E}\cdot\vec{d}\). If the displacement is exactly opposite to the field direction, the angle between \(\vec{E}\) and \(\vec{d}\) is \(180^\circ\). Hence \(\vec{E}\cdot\vec{d}=Ed\cos180^\circ=-Ed\). Substituting this gives \(V_B-V_A=-(-Ed)=+Ed\). This means potential increases when we move opposite to the electric field. The direction-related point is that potential decreases along \(\vec{E}\), so it must increase against \(\vec{E}\).
102. A uniform electric field of magnitude \(1.0\times10^3\,\text{N C}^{-1}\) acts in a region. A displacement of \(10\,\text{cm}\) is made at an angle \(60^\circ\) with the field direction. What is the potential difference between final and initial points?
ⓐ. \(+50\,\text{V}\)
ⓑ. \(-100\,\text{V}\)
ⓒ. \(-50\,\text{V}\)
ⓓ. \(+100\,\text{V}\)
Correct Answer: \(-50\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(E=1.0\times10^3\,\text{N C}^{-1}\), \(d=10\,\text{cm}=0.10\,\text{m}\), and the angle between \(\vec{E}\) and \(\vec{d}\) is \(60^\circ\).
\( \textbf{Required:} \) Potential difference \(V_B-V_A\).
\( \textbf{Uniform-field relation:} \)
\[
V_B-V_A=-\vec{E}\cdot\vec{d}
\]
\( \textbf{Dot product:} \)
\[
\vec{E}\cdot\vec{d}=Ed\cos60^\circ
\]
\( \textbf{Substitution:} \)
\[
V_B-V_A=-(1.0\times10^3)(0.10)\left(\frac{1}{2}\right)
\]
\( \textbf{Simplification:} \)
\[
(1.0\times10^3)(0.10)=100
\]
\[
100\times\frac{1}{2}=50
\]
\( \textbf{Sign result:} \)
\[
V_B-V_A=-50\,\text{V}
\]
\( \textbf{Final answer:} \) The potential difference is \(-50\,\text{V}\), because the displacement has a component along the field.
103. In a one-dimensional region, the electric potential is given by \(V=30-5x\), where \(V\) is in \(\text{V}\) and \(x\) is in \(\text{m}\). What is the electric field component \(E_x\)?
ⓐ. \(-5\,\text{V m}^{-1}\)
ⓑ. \(-30\,\text{V m}^{-1}\)
ⓒ. \(+30\,\text{V m}^{-1}\)
ⓓ. \(+5\,\text{V m}^{-1}\)
Correct Answer: \(+5\,\text{V m}^{-1}\)
Explanation: \( \textbf{Given relation:} \) \(V=30-5x\).
\( \textbf{Required:} \) Electric field component \(E_x\).
\( \textbf{Field-potential relation:} \)
\[
E_x=-\frac{dV}{dx}
\]
\( \textbf{Differentiate potential:} \)
\[
\frac{dV}{dx}=-5\,\text{V m}^{-1}
\]
\( \textbf{Apply the negative sign:} \)
\[
E_x=-(-5\,\text{V m}^{-1})
\]
\( \textbf{Simplification:} \)
\[
E_x=+5\,\text{V m}^{-1}
\]
\( \textbf{Unit equivalence:} \) \(\text{V m}^{-1}\) is equivalent to \(\text{N C}^{-1}\).
\( \textbf{Direction check:} \) Since potential decreases as \(x\) increases, the field points along \(+x\).
\( \textbf{Final answer:} \) \(E_x=+5\,\text{V m}^{-1}\).
104. A graph of \(V\) versus \(x\) is horizontal over a certain interval. What can be concluded about the electric field in that interval?
ⓐ. \(E_x\) is maximum
ⓑ. \(E_x\) is negative and constant
ⓒ. \(E_x\) is positive and constant
ⓓ. \(E_x=0\)
Correct Answer: \(E_x=0\)
Explanation: In one dimension, the electric field is related to the slope of the potential graph by \(E_x=-\frac{dV}{dx}\). A horizontal \(V\)-versus-\(x\) graph has zero slope. Therefore, \(\frac{dV}{dx}=0\). Substituting into the field relation gives \(E_x=0\). A horizontal graph means the potential is constant, not that the potential is necessarily zero. A common graph-interpretation mistake is to confuse a large value of \(V\) with a large value of \(\vec{E}\); the field depends on the gradient of \(V\), not the value of \(V\) itself.
105. Complete the statement correctly: In electrostatics, electric field is directed from ______ potential to ______ potential.
ⓐ. higher, lower
ⓑ. zero, positive
ⓒ. negative, positive only
ⓓ. lower, higher
Correct Answer: higher, lower
Explanation: The relation \(\vec{E}=-\nabla V\) shows that electric field points in the direction of decreasing potential. In one dimension, this appears as \(E_x=-\frac{dV}{dx}\). The minus sign does not mean the field is always negative; it means the field points opposite to the direction of increasing potential. A positive test charge released from rest moves in the direction of \(\vec{E}\) and tends to lose electrostatic potential energy. This rule is independent of whether the potential values themselves are positive or negative. The safe memory is that \(\vec{E}\) points downhill on the potential map.
106. Which row correctly relates the sign of \(\frac{dV}{dx}\) to the direction of \(E_x\)?
| Row | Sign of \(\frac{dV}{dx}\) | Direction of \(E_x\) |
|---|
| P | Positive | Along \(+x\) |
| Q | Positive | Along \(-x\) |
| R | Negative | Along \(-x\) |
| S | Zero | Always along \(+x\) |
ⓐ. Row P
ⓑ. Row S
ⓒ. Row R
ⓓ. Row Q
Correct Answer: Row Q
Explanation: The one-dimensional relation is \(E_x=-\frac{dV}{dx}\). If \(\frac{dV}{dx}\) is positive, then \(E_x\) is negative. A negative \(E_x\) means the electric field is directed along \(-x\). Row R is incorrect because if \(\frac{dV}{dx}\) is negative, then \(E_x\) is positive. Row S is incorrect because if \(\frac{dV}{dx}=0\), then \(E_x=0\), not along \(+x\). The sign in the formula must be applied after reading the slope of the \(V\)-versus-\(x\) graph.
107. A uniform electric field \(E=1.5\times10^3\,\text{N C}^{-1}\) is directed along \(+x\). A charge \(+2.0\,\mu\text{C}\) moves from \(x=0\) to \(x=4.0\,\text{cm}\). What are \(\Delta V\), \(\Delta U\), and work done by the field?
ⓐ. \(\Delta V=-60\,\text{V}\), \(\Delta U=+1.2\times10^{-4}\,\text{J}\), \(W_{\text{field}}=-1.2\times10^{-4}\,\text{J}\)
ⓑ. \(\Delta V=+60\,\text{V}\), \(\Delta U=+1.2\times10^{-4}\,\text{J}\), \(W_{\text{field}}=-1.2\times10^{-4}\,\text{J}\)
ⓒ. \(\Delta V=-60\,\text{V}\), \(\Delta U=-1.2\times10^{-4}\,\text{J}\), \(W_{\text{field}}=+1.2\times10^{-4}\,\text{J}\)
ⓓ. \(\Delta V=0\,\text{V}\), \(\Delta U=0\,\text{J}\), \(W_{\text{field}}=0\,\text{J}\)
Correct Answer: \(\Delta V=-60\,\text{V}\), \(\Delta U=-1.2\times10^{-4}\,\text{J}\), \(W_{\text{field}}=+1.2\times10^{-4}\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(E=1.5\times10^3\,\text{N C}^{-1}\), \(d=4.0\,\text{cm}=0.040\,\text{m}\), and \(q=+2.0\,\mu\text{C}=2.0\times10^{-6}\,\text{C}\).
\( \textbf{Direction condition:} \) The displacement is along \(+x\), the same direction as \(\vec{E}\).
\( \textbf{Potential difference:} \)
\[
\Delta V=-Ed
\]
\( \textbf{Substitution for \(\Delta V\):} \)
\[
\Delta V=-(1.5\times10^3)(0.040)
\]
\( \textbf{Potential difference result:} \)
\[
\Delta V=-60\,\text{V}
\]
\( \textbf{Potential-energy change:} \)
\[
\Delta U=q\Delta V
\]
\( \textbf{Substitution for \(\Delta U\):} \)
\[
\Delta U=(2.0\times10^{-6})(-60)=-120\times10^{-6}\,\text{J}
\]
\( \textbf{Simplification:} \)
\[
\Delta U=-1.2\times10^{-4}\,\text{J}
\]
\( \textbf{Field-work relation:} \)
\[
W_{\text{field}}=-\Delta U=+1.2\times10^{-4}\,\text{J}
\]
\( \textbf{Final answer:} \) \(\Delta V=-60\,\text{V}\), \(\Delta U=-1.2\times10^{-4}\,\text{J}\), and \(W_{\text{field}}=+1.2\times10^{-4}\,\text{J}\).
108. The electrostatic potential energy of two point charges separated by distance \(r\) is \(U=k\frac{q_1q_2}{r}\). Which statement correctly interprets the sign of \(U\)?
ⓐ. \(U\) is always zero because potential energy is measured from infinity
ⓑ. \(U\) is always positive because \(r\) is positive
ⓒ. \(U\) is positive for like charges and negative for unlike charges
ⓓ. \(U\) is positive for unlike charges and negative for like charges
Correct Answer: \(U\) is positive for like charges and negative for unlike charges
Explanation: In \(U=k\frac{q_1q_2}{r}\), the constants \(k\) and \(r\) are positive. Therefore, the sign of \(U\) is decided by the product \(q_1q_2\). For like charges, \(q_1q_2\gt0\), so \(U\) is positive. For unlike charges, \(q_1q_2\lt0\), so \(U\) is negative. The reference \(U=0\) is usually taken when the charges are infinitely separated. Positive \(U\) means external work is required to assemble the pair from infinity, while negative \(U\) means energy is released during assembly.
109. Two charges \(+2.0\,\mu\text{C}\) and \(+3.0\,\mu\text{C}\) are separated by \(0.30\,\text{m}\) in vacuum. What is their electrostatic potential energy? Take \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
ⓐ. \(+0.18\,\text{J}\)
ⓑ. \(-0.54\,\text{J}\)
ⓒ. \(-0.18\,\text{J}\)
ⓓ. \(+0.54\,\text{J}\)
Correct Answer: \(+0.18\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(q_1=+2.0\,\mu\text{C}=2.0\times10^{-6}\,\text{C}\), \(q_2=+3.0\,\mu\text{C}=3.0\times10^{-6}\,\text{C}\), \(r=0.30\,\text{m}\), and \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
\( \textbf{Required:} \) Electrostatic potential energy \(U\) of the pair.
\( \textbf{Formula:} \)
\[
U=k\frac{q_1q_2}{r}
\]
\( \textbf{Sign expectation:} \) Both charges are positive, so \(U\) should be positive.
\( \textbf{Substitution:} \)
\[
U=\frac{(9.0\times10^9)(2.0\times10^{-6})(3.0\times10^{-6})}{0.30}
\]
\( \textbf{Numerator:} \)
\[
(9.0\times10^9)(6.0\times10^{-12})=54\times10^{-3}
\]
\( \textbf{Division:} \)
\[
U=\frac{54\times10^{-3}}{0.30}=180\times10^{-3}\,\text{J}
\]
\( \textbf{Final simplification:} \)
\[
U=0.18\,\text{J}
\]
\( \textbf{Final answer:} \) The potential energy is \(+0.18\,\text{J}\).
110. Two charges \(+4.0\,\mu\text{C}\) and \(-2.0\,\mu\text{C}\) are separated by \(0.20\,\text{m}\) in vacuum. What is the electrostatic potential energy of the pair? Take \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
ⓐ. \(-0.18\,\text{J}\)
ⓑ. \(+0.36\,\text{J}\)
ⓒ. \(-0.36\,\text{J}\)
ⓓ. \(+0.18\,\text{J}\)
Correct Answer: \(-0.36\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(q_1=+4.0\times10^{-6}\,\text{C}\), \(q_2=-2.0\times10^{-6}\,\text{C}\), \(r=0.20\,\text{m}\), and \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
\( \textbf{Required:} \) Pair potential energy \(U\).
\( \textbf{Formula:} \)
\[
U=k\frac{q_1q_2}{r}
\]
\( \textbf{Sign expectation:} \) The charges are unlike, so \(q_1q_2\lt0\) and \(U\) must be negative.
\( \textbf{Substitution:} \)
\[
U=\frac{(9.0\times10^9)(+4.0\times10^{-6})(-2.0\times10^{-6})}{0.20}
\]
\( \textbf{Charge product:} \)
\[
(+4.0\times10^{-6})(-2.0\times10^{-6})=-8.0\times10^{-12}
\]
\( \textbf{Numerator:} \)
\[
(9.0\times10^9)(-8.0\times10^{-12})=-72\times10^{-3}
\]
\( \textbf{Division:} \)
\[
U=\frac{-72\times10^{-3}}{0.20}=-360\times10^{-3}\,\text{J}
\]
\( \textbf{Final answer:} \) \(U=-0.36\,\text{J}\), showing that the assembled unlike-charge pair is bound relative to infinite separation.
111. What is the work done by an external agent in slowly assembling two charges \(q_1\) and \(q_2\) from infinity to a separation \(r\)?
ⓐ. \(W_{\text{ext}}=k\frac{q_1q_2}{r}\)
ⓑ. \(W_{\text{ext}}=k\frac{|q_1+q_2|}{r^2}\)
ⓒ. \(W_{\text{ext}}=0\) for all charge pairs
ⓓ. \(W_{\text{ext}}=-k\frac{q_1q_2}{r}\)
Correct Answer: \(W_{\text{ext}}=k\frac{q_1q_2}{r}\)
Explanation: When two charges are assembled slowly from infinity, their final electrostatic potential energy is \(U=k\frac{q_1q_2}{r}\). At infinity, the reference potential energy is taken as \(0\). For slow assembly, the external work equals the change in electrostatic potential energy, so \(W_{\text{ext}}=\Delta U=U-0\). Therefore, \(W_{\text{ext}}=k\frac{q_1q_2}{r}\). For like charges this is positive, because the external agent works against repulsion. For unlike charges this is negative, because attraction helps the assembly and the external agent may need to restrain the charges.
112. A pair of like charges is brought closer from separation \(2r\) to \(r\). How does their electrostatic potential energy change?
ⓐ. It becomes zero
ⓑ. It becomes twice its previous value
ⓒ. It becomes half of its previous value
ⓓ. It becomes four times its previous value
Correct Answer: It becomes twice its previous value
Explanation: \( \textbf{Pair-energy relation:} \)
\[
U=k\frac{q_1q_2}{r}
\]
\( \textbf{For fixed charges:} \) \(U\propto\frac{1}{r}\).
\( \textbf{Initial separation:} \) \(r_i=2r\), so
\[
U_i=k\frac{q_1q_2}{2r}
\]
\( \textbf{Final separation:} \) \(r_f=r\), so
\[
U_f=k\frac{q_1q_2}{r}
\]
\( \textbf{Ratio:} \)
\[
\frac{U_f}{U_i}=\frac{k\frac{q_1q_2}{r}}{k\frac{q_1q_2}{2r}}=2
\]
\( \textbf{Sign note:} \) For like charges, \(U\) is positive, so bringing them closer increases the positive potential energy.
\( \textbf{Final answer:} \) The potential energy becomes twice its previous value.
113. A pair of unlike charges is allowed to move closer from a large separation to a smaller separation. What happens to the electrostatic potential energy of the pair?
ⓐ. It becomes less negative and approaches zero
ⓑ. It becomes more negative
ⓒ. It remains zero because the net charge may be zero
ⓓ. It becomes more positive
Correct Answer: It becomes more negative
Explanation: For unlike charges, \(q_1q_2\lt0\), so the pair potential energy \(U=k\frac{q_1q_2}{r}\) is negative. When the charges move closer, \(r\) decreases. Since the magnitude \(\left|\frac{1}{r}\right|\) increases, the magnitude of the negative potential energy increases. Therefore, \(U\) becomes more negative. Physically, attraction pulls the charges together and releases energy. The zero of potential energy is at infinite separation, so a closer unlike-charge pair lies at lower energy than the separated state.
114. Assertion: The potential energy of two unlike charges is negative when the zero of energy is taken at infinite separation.
Reason: Work is released by the electric field when unlike charges are assembled from infinity.
ⓐ. 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: For two unlike charges, the product \(q_1q_2\) is negative. Therefore, \(U=k\frac{q_1q_2}{r}\) is negative when \(U=0\) is chosen at infinity. During assembly from infinity, the attractive electrostatic force helps bring the charges closer. The electric field does positive work, so the change in electrostatic potential energy is negative because \(\Delta U=-W_{\text{field}}\). The Reason correctly explains why the final energy is below the zero-energy reference at infinity. A sign-related mistake is to think “energy released” means positive stored potential energy; it actually means the stored electrostatic potential energy decreases.
115. Two charge pairs are compared at the same separation \(r\). Pair P has charges \(+q\) and \(+3q\). Pair Q has charges \(+2q\) and \(-2q\). Which comparison of their potential energies is correct?
ⓐ. \(U_P=-\frac{3kq^2}{r}\) and \(U_Q=+\frac{4kq^2}{r}\)
ⓑ. \(U_P=0\) and \(U_Q=0\)
ⓒ. \(U_P=\frac{kq^2}{3r}\) and \(U_Q=-\frac{kq^2}{4r}\)
ⓓ. \(U_P=\frac{3kq^2}{r}\) and \(U_Q=-\frac{4kq^2}{r}\)
Correct Answer: \(U_P=\frac{3kq^2}{r}\) and \(U_Q=-\frac{4kq^2}{r}\)
Explanation: \( \textbf{Pair-energy formula:} \)
\[
U=k\frac{q_1q_2}{r}
\]
\( \textbf{Pair P charges:} \) \(q_1=+q\) and \(q_2=+3q\).
\( \textbf{Energy of Pair P:} \)
\[
U_P=k\frac{(+q)(+3q)}{r}=\frac{3kq^2}{r}
\]
\( \textbf{Pair Q charges:} \) \(q_1=+2q\) and \(q_2=-2q\).
\( \textbf{Energy of Pair Q:} \)
\[
U_Q=k\frac{(+2q)(-2q)}{r}
\]
\( \textbf{Simplification for Pair Q:} \)
\[
U_Q=-\frac{4kq^2}{r}
\]
\( \textbf{Sign interpretation:} \) Pair P is like-charge repulsive and has positive energy, while Pair Q is unlike-charge attractive and has negative energy.
\( \textbf{Final answer:} \) \(U_P=\frac{3kq^2}{r}\) and \(U_Q=-\frac{4kq^2}{r}\).
116. For a system of three point charges \(q_1\), \(q_2\), and \(q_3\), which expression correctly gives the total electrostatic potential energy?
ⓐ. \(U=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\)
ⓑ. \(U=k\left(\frac{q_1q_2}{r_{12}^2}+\frac{q_2q_3}{r_{23}^2}+\frac{q_3q_1}{r_{31}^2}\right)\)
ⓒ. \(U=2k\left(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_3q_1}{r_{31}}\right)\)
ⓓ. \(U=k\left(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_3q_1}{r_{31}}\right)\)
Correct Answer: \(U=k\left(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_3q_1}{r_{31}}\right)\)
Explanation: The electrostatic potential energy of a system of point charges is obtained by adding the potential energy of each distinct pair. For three charges, the distinct pairs are \((q_1,q_2)\), \((q_2,q_3)\), and \((q_3,q_1)\). Each pair contributes \(k\frac{q_iq_j}{r_{ij}}\). The terms must contain products of two charges because energy belongs to an interacting pair, not to a single isolated charge. The distance appears as \(r_{ij}\), not \(r_{ij}^2\), because pair potential energy varies as \(\frac{1}{r}\). The factor \(2\) is not used because each pair must be counted once only; double counting is a common error in system-energy questions.
117. Three identical charges \(+q\), \(+q\), and \(+q\) are placed at the vertices of an equilateral triangle of side \(a\). What is the total electrostatic potential energy of the system?
ⓐ. \(\frac{3kq^2}{a}\)
ⓑ. \(\frac{6kq^2}{a}\)
ⓒ. \(\frac{kq^2}{a}\)
ⓓ. \(\frac{2kq^2}{a}\)
Correct Answer: \(\frac{3kq^2}{a}\)
Explanation: \( \textbf{Given:} \) Three charges are identical, each equal to \(+q\), and every pair separation is \(a\).
\( \textbf{Required:} \) Total potential energy of the three-charge system.
\( \textbf{Pair-energy rule:} \)
\[
U_{\text{total}}=\sum_{i\lt j}k\frac{q_iq_j}{r_{ij}}
\]
\( \textbf{Number of distinct pairs:} \) Three charges make three distinct pairs.
\( \textbf{Energy of each pair:} \)
\[
U_{\text{pair}}=k\frac{(+q)(+q)}{a}=\frac{kq^2}{a}
\]
\( \textbf{Total energy:} \)
\[
U=3\left(\frac{kq^2}{a}\right)
\]
\( \textbf{Final simplification:} \)
\[
U=\frac{3kq^2}{a}
\]
\( \textbf{Sign check:} \) All pairs are like-charge pairs, so every contribution is positive.
\( \textbf{Final answer:} \) \(U=\frac{3kq^2}{a}\).
118. Three charges are placed on a straight line: \(+q\) at point \(P\), \(-2q\) at point \(Q\), and \(+q\) at point \(R\). The separation \(PQ=a\) and \(QR=a\). What is the total electrostatic potential energy of the system?
ⓐ. \(+\frac{kq^2}{2a}\)
ⓑ. \(+\frac{5kq^2}{2a}\)
ⓒ. \(-\frac{7kq^2}{2a}\)
ⓓ. \(-\frac{4kq^2}{a}\)
Correct Answer: \(-\frac{7kq^2}{2a}\)
Explanation: \( \textbf{Given:} \) \(q_P=+q\), \(q_Q=-2q\), \(q_R=+q\), \(PQ=a\), \(QR=a\), and \(PR=2a\).
\( \textbf{Required:} \) Total electrostatic potential energy.
\( \textbf{Pairwise sum:} \)
\[
U=U_{PQ}+U_{QR}+U_{PR}
\]
\( \textbf{Pair \(P,Q\):} \)
\[
U_{PQ}=k\frac{(+q)(-2q)}{a}=-\frac{2kq^2}{a}
\]
\( \textbf{Pair \(Q,R\):} \)
\[
U_{QR}=k\frac{(-2q)(+q)}{a}=-\frac{2kq^2}{a}
\]
\( \textbf{Pair \(P,R\):} \)
\[
U_{PR}=k\frac{(+q)(+q)}{2a}=+\frac{kq^2}{2a}
\]
\( \textbf{Total:} \)
\[
U=-\frac{2kq^2}{a}-\frac{2kq^2}{a}+\frac{kq^2}{2a}
\]
\( \textbf{Common denominator:} \)
\[
U=\frac{-4kq^2-4kq^2+kq^2}{2a}
\]
\( \textbf{Final answer:} \) \(U=-\frac{7kq^2}{2a}\), because the two attractive pairs dominate the one repulsive pair.
119. Three charges \(+2.0\,\mu\text{C}\), \(+2.0\,\mu\text{C}\), and \(-2.0\,\mu\text{C}\) are placed at the vertices of an equilateral triangle of side \(0.30\,\text{m}\). What is the total electrostatic potential energy of the system? Take \(k=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
ⓐ. \(-0.12\,\text{J}\)
ⓑ. \(-0.24\,\text{J}\)
ⓒ. \(+0.36\,\text{J}\)
ⓓ. \(+0.12\,\text{J}\)
Correct Answer: \(-0.12\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(q_1=+2.0\times10^{-6}\,\text{C}\), \(q_2=+2.0\times10^{-6}\,\text{C}\), \(q_3=-2.0\times10^{-6}\,\text{C}\), and each separation is \(r=0.30\,\text{m}\).
\( \textbf{Required:} \) Total potential energy \(U\).
\( \textbf{Pairwise formula:} \)
\[
U=k\left(\frac{q_1q_2}{r}+\frac{q_2q_3}{r}+\frac{q_3q_1}{r}\right)
\]
\( \textbf{Equal-distance factorisation:} \)
\[
U=\frac{k}{r}(q_1q_2+q_2q_3+q_3q_1)
\]
\( \textbf{Pair products:} \)
\[
q_1q_2=+4.0\times10^{-12}\,\text{C}^2
\]
\[
q_2q_3=-4.0\times10^{-12}\,\text{C}^2
\]
\[
q_3q_1=-4.0\times10^{-12}\,\text{C}^2
\]
\( \textbf{Sum of products:} \)
\[
4.0\times10^{-12}-4.0\times10^{-12}-4.0\times10^{-12}=-4.0\times10^{-12}
\]
\( \textbf{Substitution:} \)
\[
U=\frac{(9.0\times10^9)(-4.0\times10^{-12})}{0.30}
\]
\( \textbf{Calculation:} \)
\[
U=\frac{-36\times10^{-3}}{0.30}=-0.12\,\text{J}
\]
\( \textbf{Final answer:} \) \(U=-0.12\,\text{J}\), because there are two unlike attractive pairs and only one like repulsive pair.
120. Which statement correctly explains why the total potential energy of a system of point charges is written as \(\sum_{i\lt j}k\frac{q_iq_j}{r_{ij}}\)?
ⓐ. The condition \(i\lt j\) changes potential energy into electric potential
ⓑ. The condition \(i\lt j\) makes all charge products positive
ⓒ. The condition \(i\lt j\) removes the need to know distances between charges
ⓓ. The condition \(i\lt j\) ensures that each pair is counted once
Correct Answer: The condition \(i\lt j\) ensures that each pair is counted once
Explanation: Electrostatic potential energy of a system is built from pair interactions. If one adds over all \(i\) and \(j\) freely, the pair \((q_1,q_2)\) and the pair \((q_2,q_1)\) would both be counted, even though they represent the same physical interaction. The notation \(i\lt j\) selects each distinct pair only once. It does not change the signs of the products \(q_iq_j\), so attractive and repulsive contributions remain properly signed. Distances \(r_{ij}\) are still essential because each pair energy depends on separation. The common mistake is to include both \(U_{12}\) and \(U_{21}\), which doubles the true energy.
