101. A negatively charged electroscope shows increased leaf divergence when an unknown rod is brought near its cap without touching. What is the sign of the unknown rod?
ⓐ. Positive
ⓑ. Neutral
ⓒ. Negative
ⓓ. Zero only
Correct Answer: Negative
Explanation: A negatively charged electroscope has leaves that already repel because both leaves carry negative charge. When a negatively charged rod is brought near the cap, it repels electrons in the electroscope downward toward the leaves. This increases the excess negative charge on the leaves. Greater like-charge repulsion makes the leaves diverge more. If the rod were positive, it would attract electrons upward toward the cap and reduce negative charge on the leaves. The increase in divergence therefore indicates that the unknown rod has the same sign as the electroscope. Hence, the unknown rod is negative.
102. A gold-leaf electroscope has diverged leaves. Which action most directly discharges it?
ⓐ. Bringing an insulating rod near the cap without touching
ⓑ. Placing the electroscope on a dry glass slab
ⓒ. Covering the electroscope with a plastic sheet
ⓓ. Touching the metal cap with a finger
Correct Answer: Touching the metal cap with a finger
Explanation: Diverged leaves indicate that the leaves carry like charges and repel each other. To discharge the electroscope, a path must be provided for charge transfer. Touching the metal cap with a finger can connect the electroscope to earth through the human body. If the electroscope has excess electrons, they may flow to earth. If it has a deficiency of electrons, electrons may come from earth. In either case, the charge on the leaves reduces and their repulsion decreases. Bringing an insulating object near without touching may change divergence temporarily but does not necessarily discharge the electroscope.
103. A charged rod is brought near the cap of a neutral electroscope without contact, and the leaves diverge. What does this observation show most directly?
ⓐ. Charges inside the electroscope have redistributed.
ⓑ. The electroscope has gained net charge from the rod.
ⓒ. The leaves have become oppositely charged.
ⓓ. The rod has become completely neutral.
Correct Answer: Charges inside the electroscope have redistributed.
Explanation: When a charged rod is brought near a neutral electroscope without touching it, no direct charge transfer from the rod is required. The electric influence of the rod causes mobile electrons in the conducting parts of the electroscope to shift. This produces induced charge separation between the cap and the leaves. The two leaves acquire like charges, so they repel and diverge. The electroscope as a whole can still remain neutral because no charge has entered or left it. The rod does not need to lose its charge for this effect to occur. Therefore, leaf divergence in this case is direct evidence of charge redistribution by induction.
104. A positively charged electroscope has a certain leaf divergence. When an unknown rod is brought near the cap, the divergence decreases. What is the sign of the rod?
ⓐ. Positive
ⓑ. Negative
ⓒ. Same as the electroscope
ⓓ. No charge possible
Correct Answer: Negative
Explanation: A positively charged electroscope has leaves that repel because both leaves are positively charged. A negatively charged rod near the cap repels electrons downward toward the leaves. These electrons partially neutralize the positive charge on the leaves. The repulsive force between the leaves becomes weaker. As a result, the leaf divergence decreases. A positive rod would pull electrons toward the cap and leave the leaves more positive, increasing divergence. Therefore, a decrease in divergence for a positively charged electroscope indicates a negative rod.
105. Which statement about using an electroscope to compare charge amount is most accurate?
ⓐ. Equal divergence always means equal charge for all objects in all conditions.
ⓑ. Leaf divergence gives exact charge in \(\text{C}\) without calibration.
ⓒ. Greater divergence usually means greater like-charge repulsion.
ⓓ. Leaf divergence depends only on mass and not on electric charge.
Correct Answer: Greater divergence usually means greater like-charge repulsion.
Explanation: A gold-leaf electroscope works because like charges on its leaves repel. If the charge on the leaves is greater, the repulsive force between them is generally greater. Under similar conditions, greater divergence can therefore indicate a larger charge effect. However, an ordinary electroscope is not automatically a precision charge meter. Exact measurement in \(\text{C}\) would require calibration and controlled geometry. Equal divergence in different setups does not always guarantee equal charge. The divergence is caused by electrostatic repulsion, not by the mass of the leaves alone.
106. A neutral gold-leaf electroscope is touched with a negatively charged rod and then the rod is removed. What is the final state of the leaves?
ⓐ. They remain closed because charge cannot enter metal.
ⓑ. They become oppositely charged and attract.
ⓒ. They lose all electrons and become positive.
ⓓ. They diverge because both leaves become negative.
Correct Answer: They diverge because both leaves become negative.
Explanation: Touching the cap with a negatively charged rod allows charge transfer by contact. Excess electrons can move from the rod to the conducting parts of the electroscope. These electrons spread through the cap, stem, and leaves. The leaves then acquire the same kind of charge, namely negative charge. Like charges repel, so the leaves diverge. Removing the rod afterward does not necessarily remove the charge already transferred to the electroscope. Therefore, the leaves remain separated because they carry like negative charges.
107. A gold-leaf electroscope is charged and its leaves are diverged. The cap is momentarily connected to earth. What happens to the leaves after effective discharge?
ⓐ. They diverge more because earth supplies only positive charge.
ⓑ. They collapse as like charge on the leaves is reduced.
ⓒ. They rotate around the stem because charge has direction.
ⓓ. They become permanently oppositely charged.
Correct Answer: They collapse as like charge on the leaves is reduced.
Explanation: The leaves of a charged electroscope diverge because they carry like charges. Earthing the cap provides a conducting path between the electroscope and earth. Charge can then move until the excess charge on the electroscope is removed or balanced. With much less like charge on the leaves, the repulsive force between them decreases. The leaves then collapse toward each other. Earth does not supply only positive charge; electron flow direction depends on the sign of the electroscope. The collapse indicates discharge, not conversion into oppositely charged leaves.
108. A student observes that the leaves of an electroscope diverge when a charged body is touched to its cap. Which inference is most justified?
ⓐ. The electroscope acquired charge by contact.
ⓑ. The electroscope has become magnetized permanently.
ⓒ. The leaves have acquired unequal masses.
ⓓ. The charged body must have been neutralized fully.
Correct Answer: The electroscope acquired charge by contact.
Explanation: Touching a charged body to the cap of an electroscope allows charge transfer. Since the electroscope is conducting, the transferred charge can spread to the leaves. The leaves then carry charges of the same sign. Like-charge repulsion causes them to diverge. The observation is therefore consistent with charging by contact. Magnetization is not needed to explain the leaf separation. The charged body may lose some charge during contact, but it need not become completely neutral. The most direct inference is that the electroscope has acquired charge.
109. Charges \(+3\,\mu\text{C}\), \(-5\,\mu\text{C}\), \(+8\,\mu\text{C}\), and \(-2\,\mu\text{C}\) are placed on separate small bodies in an isolated system. What is the net charge of the system?
ⓐ. \(+18\,\mu\text{C}\)
ⓑ. \(-12\,\mu\text{C}\)
ⓒ. \(-4\,\mu\text{C}\)
ⓓ. \(+4\,\mu\text{C}\)
Correct Answer: \(+4\,\mu\text{C}\)
Explanation: \( \textbf{Known charges:} \) The charges are \(+3\,\mu\text{C}\), \(-5\,\mu\text{C}\), \(+8\,\mu\text{C}\), and \(-2\,\mu\text{C}\).
\( \textbf{Required quantity:} \) Find the algebraic net charge of the system.
\( \textbf{Useful relation:} \) \(Q_{\text{net}}=q_1+q_2+q_3+q_4\).
\( \textbf{Substitution:} \) \(Q_{\text{net}}=(+3-5+8-2)\,\mu\text{C}\).
\( \textbf{Positive contribution:} \) \(+3\,\mu\text{C}+8\,\mu\text{C}=+11\,\mu\text{C}\).
\( \textbf{Negative contribution:} \) \(-5\,\mu\text{C}-2\,\mu\text{C}=-7\,\mu\text{C}\).
\( \textbf{Final simplification:} \) \(Q_{\text{net}}=+11\,\mu\text{C}-7\,\mu\text{C}=+4\,\mu\text{C}\).
\( \textbf{Final result:} \) The net charge is \(+4\,\mu\text{C}\).
110. Which expression correctly represents the additivity of electric charge for several charges?
ⓐ. \(Q_{\text{net}}=q_1q_2q_3\cdots\)
ⓑ. \(Q_{\text{net}}=\frac{q_1+q_2+q_3+\cdots}{r^2}\)
ⓒ. \(Q_{\text{net}}=q_1+q_2+q_3+\cdots\)
ⓓ. \(Q_{\text{net}}=\sqrt{q_1^2+q_2^2+q_3^2+\cdots}\)
Correct Answer: \(Q_{\text{net}}=q_1+q_2+q_3+\cdots\)
Explanation: \( \textbf{Concept involved:} \) Electric charge is additive.
\( \textbf{Meaning of additivity:} \) The total charge of a system is found by algebraically adding individual charges.
\( \textbf{Sign use:} \) Positive charges are added as positive terms, and negative charges are added as negative terms.
\( \textbf{Correct relation:} \) \(Q_{\text{net}}=q_1+q_2+q_3+\cdots\).
\( \textbf{Why not vector addition:} \) Charge is a scalar quantity with sign, not a vector requiring component addition.
\( \textbf{Why no distance factor:} \) Distance affects electrostatic force, not the algebraic total charge.
\( \textbf{Final result:} \) The additive relation for charge is \(Q_{\text{net}}=q_1+q_2+q_3+\cdots\).
111. A system contains charges \(+12\,\text{nC}\), \(-7\,\text{nC}\), and \(-5\,\text{nC}\). What is the net charge?
ⓐ. \(+24\,\text{nC}\)
ⓑ. \(0\,\text{nC}\)
ⓒ. \(-14\,\text{nC}\)
ⓓ. \(+10\,\text{nC}\)
Correct Answer: \(0\,\text{nC}\)
Explanation: \( \textbf{Given charges:} \) \(q_1=+12\,\text{nC}\), \(q_2=-7\,\text{nC}\), and \(q_3=-5\,\text{nC}\).
\( \textbf{Required:} \) Net charge \(Q_{\text{net}}\).
\( \textbf{Additivity relation:} \) \(Q_{\text{net}}=q_1+q_2+q_3\).
\( \textbf{Substitution:} \) \(Q_{\text{net}}=(+12-7-5)\,\text{nC}\).
\( \textbf{Combine negative charges:} \) \(-7\,\text{nC}-5\,\text{nC}=-12\,\text{nC}\).
\( \textbf{Cancellation:} \) \(+12\,\text{nC}+(-12\,\text{nC})=0\,\text{nC}\).
\( \textbf{Final result:} \) The net charge of the system is \(0\,\text{nC}\).
112. A body has charges \(+4\,\mu\text{C}\), \(-9\,\mu\text{C}\), and \(+2\,\mu\text{C}\) distributed on different parts of it. What is the total charge on the body?
ⓐ. \(-3\,\mu\text{C}\)
ⓑ. \(+15\,\mu\text{C}\)
ⓒ. \(+7\,\mu\text{C}\)
ⓓ. \(-15\,\mu\text{C}\)
Correct Answer: \(-3\,\mu\text{C}\)
Explanation: \( \textbf{Charges present:} \) \(+4\,\mu\text{C}\), \(-9\,\mu\text{C}\), and \(+2\,\mu\text{C}\).
\( \textbf{Quantity to find:} \) The total charge on the body.
\( \textbf{Addition rule:} \) Since charge is scalar with sign, use algebraic addition.
\( \textbf{Relation:} \) \(Q_{\text{net}}=+4\,\mu\text{C}-9\,\mu\text{C}+2\,\mu\text{C}\).
\( \textbf{Positive total:} \) \(+4\,\mu\text{C}+2\,\mu\text{C}=+6\,\mu\text{C}\).
\( \textbf{Combine with negative charge:} \) \(+6\,\mu\text{C}-9\,\mu\text{C}=-3\,\mu\text{C}\).
\( \textbf{Sign interpretation:} \) The body has excess negative charge overall.
\( \textbf{Final result:} \) The total charge is \(-3\,\mu\text{C}\).
113. Why is ordinary algebraic addition used for charges such as \(+q\) and \(-2q\)?
ⓐ. Charge has direction along the line joining charged bodies.
ⓑ. Charge always behaves like force during addition.
ⓒ. Charge is a scalar quantity with positive or negative sign.
ⓓ. Charge must be squared before addition.
Correct Answer: Charge is a scalar quantity with positive or negative sign.
Explanation: Electric charge is a scalar physical quantity, not a vector. It may have a positive or negative sign, but that sign does not represent a spatial direction. When several charges are present in a system, their total is obtained by algebraic addition. Positive charges contribute positive terms, and negative charges contribute negative terms. For example, \(+q+(-2q)=-q\). Vector addition is needed for quantities like force and electric field, not for charge itself. Therefore, signed scalar addition is the proper method for adding charges.
114. A system contains four charges. Their net charge is \(+6\,\mu\text{C}\). Three charges are \(+10\,\mu\text{C}\), \(-4\,\mu\text{C}\), and \(-8\,\mu\text{C}\). What is the fourth charge?
ⓐ. \(-8\,\mu\text{C}\)
ⓑ. \(+4\,\mu\text{C}\)
ⓒ. \(+6\,\mu\text{C}\)
ⓓ. \(+8\,\mu\text{C}\)
Correct Answer: \(+8\,\mu\text{C}\)
Explanation: \( \textbf{Given net charge:} \) \(Q_{\text{net}}=+6\,\mu\text{C}\).
\( \textbf{Known charges:} \) \(+10\,\mu\text{C}\), \(-4\,\mu\text{C}\), and \(-8\,\mu\text{C}\).
\( \textbf{Let unknown charge be:} \) \(q\).
\( \textbf{Additivity relation:} \) \(Q_{\text{net}}=+10\,\mu\text{C}-4\,\mu\text{C}-8\,\mu\text{C}+q\).
\( \textbf{Known-charge sum:} \) \(+10\,\mu\text{C}-4\,\mu\text{C}-8\,\mu\text{C}=-2\,\mu\text{C}\).
\( \textbf{Equation:} \) \(+6\,\mu\text{C}=-2\,\mu\text{C}+q\).
\( \textbf{Solving:} \) \(q=+8\,\mu\text{C}\).
\( \textbf{Final result:} \) The fourth charge is \(+8\,\mu\text{C}\).
115. Two charges \(+7\,\text{nC}\) and \(-7\,\text{nC}\) are placed in the same isolated system. Which statement about the net charge is correct?
ⓐ. The net charge is \(0\,\text{nC}\).
ⓑ. The net charge is \(+14\,\text{nC}\) because magnitudes must be added.
ⓒ. The net charge is \(-14\,\text{nC}\) because negative charge dominates.
ⓓ. The net charge is \(49\,\text{nC}\) because charges multiply.
Correct Answer: The net charge is \(0\,\text{nC}\).
Explanation: \( \textbf{Given charges:} \) \(q_1=+7\,\text{nC}\) and \(q_2=-7\,\text{nC}\).
\( \textbf{Required quantity:} \) Net charge of the system.
\( \textbf{Charge addition rule:} \) Total charge is the algebraic sum of individual charges.
\( \textbf{Substitution:} \) \(Q_{\text{net}}=q_1+q_2=+7\,\text{nC}+(-7\,\text{nC})\).
\( \textbf{Cancellation:} \) Equal positive and negative charges cancel in the algebraic sum.
\( \textbf{Simplification:} \) \(Q_{\text{net}}=0\,\text{nC}\).
\( \textbf{Meaning:} \) The system is electrically neutral overall, though it contains charged parts.
\( \textbf{Final result:} \) The net charge is \(0\,\text{nC}\).
116. A group of charges consists of \(+2\,\mu\text{C}\), \(+5\,\mu\text{C}\), \(-3\,\mu\text{C}\), and an unknown charge \(q\). If the net charge is \(+1\,\mu\text{C}\), what is \(q\)?
ⓐ. \(+1\,\mu\text{C}\)
ⓑ. \(-1\,\mu\text{C}\)
ⓒ. \(-3\,\mu\text{C}\)
ⓓ. \(+3\,\mu\text{C}\)
Correct Answer: \(-3\,\mu\text{C}\)
Explanation: \( \textbf{Given charges:} \) The charges are \(+2\,\mu\text{C}\), \(+5\,\mu\text{C}\), \(-3\,\mu\text{C}\), and \(q\).
\( \textbf{Net charge:} \) \(Q_{\text{net}}=+1\,\mu\text{C}\).
\( \textbf{Additivity rule:} \) \(Q_{\text{net}}=q_1+q_2+q_3+q\).
\( \textbf{Substitution:} \) \(+1\,\mu\text{C}=+2\,\mu\text{C}+5\,\mu\text{C}-3\,\mu\text{C}+q\).
\( \textbf{Known-charge sum:} \) \(+2\,\mu\text{C}+5\,\mu\text{C}-3\,\mu\text{C}=+4\,\mu\text{C}\).
\( \textbf{Equation:} \) \(+1\,\mu\text{C}=+4\,\mu\text{C}+q\).
\( \textbf{Solving:} \) \(q=-3\,\mu\text{C}\).
\( \textbf{Final result:} \) The unknown charge is \(-3\,\mu\text{C}\).
117. A set of charges is written as \(+4\,\text{nC}\), \(-6\,\text{nC}\), \(+2\,\text{nC}\), and \(-1\,\text{nC}\). Which common mistake gives the wrong result \(13\,\text{nC}\)?
ⓐ. Adding only the negative charges algebraically
ⓑ. Adding magnitudes and ignoring signs
ⓒ. Multiplying all signed charge values
ⓓ. Dividing the algebraic sum by \(4\)
Correct Answer: Adding magnitudes and ignoring signs
Explanation: The net charge must be found by algebraic addition. The signs \(+\) and \(-\) must be kept because they represent opposite kinds of charge. For the given charges, the correct sum is \(+4\,\text{nC}-6\,\text{nC}+2\,\text{nC}-1\,\text{nC}\). This gives \(-1\,\text{nC}\). The result \(13\,\text{nC}\) comes from adding only magnitudes: \(4+6+2+1=13\). That method finds the total amount of charge magnitude listed, not the net charge. Net charge requires cancellation between positive and negative contributions.
118. Four charged particles have charges \(+3q\), \(-q\), \(-4q\), and \(+2q\). What is the net charge of the group?
ⓐ. \(+10q\)
ⓑ. \(-10q\)
ⓒ. \(+2q\)
ⓓ. \(0\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given charges:} \) \(+3q\), \(-q\), \(-4q\), and \(+2q\).
\( \textbf{Required quantity:} \) Net charge of the group.
\( \textbf{Charge addition:} \) \(Q_{\text{net}}=+3q-q-4q+2q\).
\( \textbf{Positive terms:} \) \(+3q+2q=+5q\).
\( \textbf{Negative terms:} \) \(-q-4q=-5q\).
\( \textbf{Cancellation:} \) \(+5q+(-5q)=0\).
\( \textbf{Meaning:} \) The group is neutral overall, even though it contains charged particles.
\( \textbf{Final result:} \) \(Q_{\text{net}}=0\).
119. Which statement about a system having net charge \(0\) is most accurate?
ⓐ. It may contain charges, but their algebraic sum is zero.
ⓑ. It cannot contain any positive or negative charge at all.
ⓒ. It must contain only positive charges of equal magnitude.
ⓓ. It must contain only negative charges of equal magnitude.
Correct Answer: It may contain charges, but their algebraic sum is zero.
Explanation: A system with net charge \(0\) is electrically neutral overall. This does not necessarily mean that no charged particles or charged parts are present. Equal positive and negative contributions can cancel in the algebraic sum. For example, \(+5\,\mu\text{C}\) and \(-5\,\mu\text{C}\) together give a net charge of \(0\,\mu\text{C}\). The individual charges still exist and may be separated in space. Neutrality is a statement about the total charge, not about the absence of charge everywhere. Therefore, a neutral system may contain charges whose signed sum is zero.
120. A glass rod and a silk cloth are initially neutral. After rubbing, the glass rod has charge \(+8\,\text{nC}\). If the rod-cloth pair is isolated, what is the total final charge of the pair?
ⓐ. \(+8\,\text{nC}\)
ⓑ. \(-8\,\text{nC}\)
ⓒ. \(0\,\text{nC}\)
ⓓ. \(+16\,\text{nC}\)
Correct Answer: \(0\,\text{nC}\)
Explanation: \( \textbf{Initial state:} \) The rod and cloth are initially neutral, so \(Q_{\text{initial}}=0\,\text{nC}\).
\( \textbf{Given after rubbing:} \) The rod has charge \(+8\,\text{nC}\).
\( \textbf{Conservation principle:} \) In an isolated system, total charge remains constant.
\( \textbf{Charge on cloth:} \) To keep the total zero, the cloth must have \(-8\,\text{nC}\).
\( \textbf{Final total:} \) \(Q_{\text{final}}=+8\,\text{nC}+(-8\,\text{nC})\).
\( \textbf{Simplification:} \) \(Q_{\text{final}}=0\,\text{nC}\).
\( \textbf{Physical meaning:} \) Rubbing transfers electrons; it does not create net charge in the isolated pair.
\( \textbf{Final result:} \) The total final charge of the pair is \(0\,\text{nC}\).
