101. Study the statements about conservation of charge.
I. The total charge of an isolated system remains constant.
II. Charge can be transferred from one body to another.
III. Ordinary charging means net charge is created from nothing.
IV. Pair production can preserve total charge if equal and opposite charges appear.
The supported statements are
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is the basic conservation law for electric charge. Statement II is valid because charging by friction, contact, and induction can involve transfer or flow of charge. Statement IV is valid because equal and opposite charges have zero algebraic total. Statement III is not valid for ordinary charging because the process redistributes charge rather than creating net charge from nothing. A body may gain charge, but another body or the surroundings must account for the balancing change. The conservation law must be applied to the complete isolated system.
102. Two isolated conducting spheres have total charge \(+14\,\mu\text{C}\). After they touch and separate, one sphere has charge \(+9\,\mu\text{C}\). The charge on the other sphere is
ⓐ. \(-23\,\mu\text{C}\)
ⓑ. \(-5\,\mu\text{C}\)
ⓒ. \(+23\,\mu\text{C}\)
ⓓ. \(+5\,\mu\text{C}\)
Correct Answer: \(+5\,\mu\text{C}\)
Explanation: \( \textbf{Total charge of isolated system:} \) \(Q_{\text{total}}=+14\,\mu\text{C}\).
\( \textbf{Final charge on one sphere:} \) \(q_1=+9\,\mu\text{C}\).
\( \textbf{Let the charge on the other sphere be:} \) \(q_2\).
\( \textbf{Conservation relation:} \)
\[
q_1+q_2=Q_{\text{total}}
\]
\( \textbf{Substitution:} \)
\[
+9\,\mu\text{C}+q_2=+14\,\mu\text{C}
\]
\( \textbf{Solving:} \)
\[
q_2=+14\,\mu\text{C}-9\,\mu\text{C}
\]
\( \textbf{Result:} \)
\[
q_2=+5\,\mu\text{C}
\]
\( \textbf{Check:} \) \(+9\,\mu\text{C}+5\,\mu\text{C}=+14\,\mu\text{C}\), so the total is conserved.
\( \textbf{Final answer:} \) The charge on the other sphere is \(+5\,\mu\text{C}\).
103. A body becomes negatively charged after receiving electrons from another body. Conservation of charge requires that the other body
ⓐ. also gains the same number of electrons
ⓑ. loses protons and becomes neutral in every case
ⓒ. loses electrons and becomes relatively positive
ⓓ. creates negative charge inside itself
Correct Answer: loses electrons and becomes relatively positive
Explanation: If one body receives electrons, it gains negative charge. Those electrons must come from another body or from the surroundings. The body that supplies electrons loses negative charge and becomes relatively positive. Conservation of charge connects the gain of one body with the loss of another. Protons are not normally transferred between ordinary solid bodies during such charging. The charge imbalance is produced by electron transfer, not by creating extra negative charge inside one body alone.
104. A charge conservation record is shown below.
| Stage | Charge on body P | Charge on body Q | Total charge |
|---|
| Before transfer | \(+6\,\text{nC}\) | \(-2\,\text{nC}\) | \(+4\,\text{nC}\) |
| After transfer | \(+1\,\text{nC}\) | ? | \(+4\,\text{nC}\) |
The missing charge on body Q after transfer is
ⓐ. \(-3\,\text{nC}\)
ⓑ. \(+3\,\text{nC}\)
ⓒ. \(+5\,\text{nC}\)
ⓓ. \(-5\,\text{nC}\)
Correct Answer: \(+3\,\text{nC}\)
Explanation: \( \textbf{Total charge before transfer:} \) \(+4\,\text{nC}\).
\( \textbf{Conservation condition:} \) The total charge after transfer must remain \(+4\,\text{nC}\).
\( \textbf{After-transfer charge on P:} \) \(+1\,\text{nC}\).
\( \textbf{Let after-transfer charge on Q be:} \) \(q_Q\).
\( \textbf{Total-charge equation:} \)
\[
(+1\,\text{nC})+q_Q=+4\,\text{nC}
\]
\( \textbf{Solving:} \)
\[
q_Q=+4\,\text{nC}-1\,\text{nC}
\]
\( \textbf{Result:} \)
\[
q_Q=+3\,\text{nC}
\]
\( \textbf{Meaning:} \) Body Q has changed from \(-2\,\text{nC}\) to \(+3\,\text{nC}\), but the two-body total is unchanged.
\( \textbf{Final answer:} \) The missing charge is \(+3\,\text{nC}\).
105. The statement \(q=ne\) represents the quantisation of electric charge, where \(n\) must be
ⓐ. any decimal number
ⓑ. any fraction less than \(1\)
ⓒ. only a positive even number
ⓓ. an integer
Correct Answer: an integer
Explanation: Quantisation of charge means that the charge on an isolated body occurs in integral multiples of the elementary charge \(e\). The relation is \(q=ne\), where \(n\in\mathbb{Z}\). The integer \(n\) may be positive, negative, or zero depending on the net charge of the body. A fractional value such as \(0.5e\) is not allowed as the net charge of an isolated body in elementary electrostatics. The sign of \(n\) tells whether the body has a net positive or negative charge. The condition \(n\in\mathbb{Z}\) is the central point, not merely the formula shape.
106. A body has charge \(+4e\). This means the body has
ⓐ. a net charge of four positive elementary units
ⓑ. a vector charge pointing four units to the right
ⓒ. a charge smaller than one elementary charge
ⓓ. no possible relation with elementary charge
Correct Answer: a net charge of four positive elementary units
Explanation: A charge written as \(+4e\) means four times the elementary charge with positive sign. It is an allowed charge because it is an integral multiple of \(e\). The positive sign tells the kind of net charge, not a spatial direction. The charge is not smaller than \(e\); its magnitude is \(4e\). The expression directly uses the quantisation relation \(q=ne\). In this notation, the integer multiplier \(4\) carries the allowed count of elementary charge units.
107. The elementary charge is commonly taken as
ⓐ. \(1.602\times10^{-19}\,\text{C}\)
ⓑ. \(1.602\times10^{-20}\,\text{C}\)
ⓒ. \(9.0\times10^9\,\text{C}\)
ⓓ. \(8.85\times10^{-12}\,\text{C}\)
Correct Answer: \(1.602\times10^{-19}\,\text{C}\)
Explanation: The elementary charge has magnitude \(e=1.602\times10^{-19}\,\text{C}\). The charge of a proton is \(+e\), while the charge of an electron is \(-e\). The value \(9.0\times10^9\) is associated with the Coulomb constant \(k\), not with elementary charge. The value \(8.85\times10^{-12}\) is associated with the vacuum permittivity \(\varepsilon_0\), not with \(e\). The very small value of \(e\) explains why charge often appears continuous in macroscopic situations. The unit remains the coulomb \(\text{C}\).
108. An isolated oil drop has charge \(q=-6.4\times10^{-19}\,\text{C}\). Taking \(e=1.6\times10^{-19}\,\text{C}\), the value of \(n\) in \(q=ne\) is
ⓐ. \(+4\)
ⓑ. \(+0.25\)
ⓒ. \(-4\)
ⓓ. \(-2\)
Correct Answer: \(-4\)
Explanation: \( \textbf{Given charge:} \) \(q=-6.4\times10^{-19}\,\text{C}\).
\( \textbf{Elementary charge magnitude:} \) \(e=1.6\times10^{-19}\,\text{C}\).
\( \textbf{Quantisation relation:} \)
\[
q=ne
\]
\( \textbf{Required value:} \) \(n=\frac{q}{e}\).
\( \textbf{Substitution with sign:} \)
\[
n=\frac{-6.4\times10^{-19}}{1.6\times10^{-19}}
\]
\( \textbf{Numerical part:} \)
\[
\frac{-6.4}{1.6}=-4
\]
\( \textbf{Power of ten cancellation:} \)
\[
10^{-19}/10^{-19}=1
\]
\( \textbf{Result:} \)
\[
n=-4
\]
\( \textbf{Sign meaning:} \) The negative value shows an excess of four electrons in net.
\( \textbf{Final answer:} \) The value of \(n\) is \(-4\).
109. A listed charge value for an isolated particle-like body is \(2.4\times10^{-19}\,\text{C}\). With \(e=1.6\times10^{-19}\,\text{C}\), this value is not allowed as a net charge because
ⓐ. it gives \(n=2\), which is too large
ⓑ. it gives \(n=1.5\), not an integer
ⓒ. it has the unit \(\text{C}\)
ⓓ. it is positive
Correct Answer: it gives \(n=1.5\), not an integer
Explanation: \( \textbf{Given charge:} \) \(q=2.4\times10^{-19}\,\text{C}\).
\( \textbf{Elementary charge:} \) \(e=1.6\times10^{-19}\,\text{C}\).
\( \textbf{Quantisation check:} \)
\[
n=\frac{q}{e}
\]
\( \textbf{Substitution:} \)
\[
n=\frac{2.4\times10^{-19}}{1.6\times10^{-19}}
\]
\( \textbf{Calculation:} \)
\[
n=1.5
\]
\( \textbf{Condition for isolated net charge:} \) \(n\) must be an integer.
\( \textbf{Decision:} \) \(1.5\notin\mathbb{Z}\), so the charge is not an allowed isolated net charge.
\( \textbf{Unit note:} \) Having the unit \(\text{C}\) is necessary but not enough; the value must also be an integral multiple of \(e\).
\( \textbf{Final answer:} \) The charge is not allowed because \(n=1.5\).
110. A proton and an electron have charges
ⓐ. proton \(+e\), electron \(-e\)
ⓑ. proton \(-e\), electron \(+e\)
ⓒ. proton \(0\), electron \(+e\)
ⓓ. proton \(+2e\), electron \(-2e\)
Correct Answer: proton \(+e\), electron \(-e\)
Explanation: A proton carries positive elementary charge \(+e\). An electron carries negative elementary charge \(-e\). Their charge magnitudes are equal, but their signs are opposite. This equality of magnitude is why an atom with equal numbers of protons and electrons can be neutral. The signs do not describe spatial directions; they describe opposite kinds of electric charge. This proton-electron sign convention is the basis for interpreting electron gain and electron loss during charging.
111. A macroscopic object may appear to have continuously variable charge because
ⓐ. charge is not quantised for large bodies
ⓑ. \(e\) is very small compared with ordinary charges
ⓒ. protons and electrons have no charge in large bodies
ⓓ. the unit \(\text{C}\) disappears at large scale
Correct Answer: \(e\) is very small compared with ordinary charges
Explanation: Electric charge is always quantised in the relation \(q=ne\). For macroscopic objects, the number \(n\) can be enormously large. Since \(e=1.602\times10^{-19}\,\text{C}\) is very small, a change by one elementary charge is usually too tiny to notice in ordinary measurements. This makes charge appear continuous at large scales. The quantisation rule itself does not stop applying to large bodies. The apparent continuity is due to measurement scale, not due to loss of quantisation.
112. Study the table and identify the row that violates charge quantisation for an isolated body. Take \(e\) as the elementary charge.
| Row | Charge written | Value of \(n\) in \(q=ne\) |
|---|
| P | \(+7e\) | \(+7\) |
| Q | \(-3e\) | \(-3\) |
| R | \(0\) | \(0\) |
| S | \(+\frac{2}{3}e\) | \(+\frac{2}{3}\) |
ⓐ. Row P
ⓑ. Row S
ⓒ. Row R
ⓓ. Row Q
Correct Answer: Row S
Explanation: Quantisation requires \(q=ne\), where \(n\) must be an integer. Rows P and Q have integer values \(+7\) and \(-3\), so they are allowed charge values. Row R is also allowed because \(n=0\) represents zero net charge. Row S has \(n=+\frac{2}{3}\), which is not an integer. Therefore, row S violates the quantisation rule for an isolated net charge. The fraction is the problem, not the positive sign.
113. Assertion: A body cannot have net charge \(0.5e\) as an isolated charge.
Reason: The net charge of an isolated body must be an integral multiple of \(e\).
ⓐ. 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: The Assertion is true because \(0.5e\) would correspond to \(n=0.5\) in the relation \(q=ne\). The Reason is also true because charge quantisation requires \(n\in\mathbb{Z}\). Since \(0.5\) is not an integer, \(0.5e\) is not an allowed isolated net charge in elementary electrostatics. Values such as \(0\), \(+e\), \(-2e\), and \(+5e\) are allowed because their multipliers are integers. The Reason directly explains the rejection of \(0.5e\). The allowed values form discrete steps of size \(e\).
114. A tiny body has a deficiency of \(2.0\times10^6\) electrons. Its net charge is
ⓐ. \(-1.25\times10^{25}\,\text{C}\)
ⓑ. \(-3.2\times10^{-13}\,\text{C}\)
ⓒ. \(+1.25\times10^{25}\,\text{C}\)
ⓓ. \(+3.2\times10^{-13}\,\text{C}\)
Correct Answer: \(+3.2\times10^{-13}\,\text{C}\)
Explanation: \( \textbf{Electron deficiency:} \) \(n=2.0\times10^6\) electrons are missing.
\( \textbf{Elementary charge:} \) \(e=1.6\times10^{-19}\,\text{C}\).
\( \textbf{Sign decision:} \) A deficiency of electrons gives positive charge.
\( \textbf{Magnitude relation:} \)
\[
q=ne
\]
\( \textbf{Substitution:} \)
\[
q=(2.0\times10^6)(1.6\times10^{-19}\,\text{C})
\]
\( \textbf{Numerical product:} \)
\[
2.0\times1.6=3.2
\]
\( \textbf{Power of ten:} \)
\[
10^6\times10^{-19}=10^{-13}
\]
\( \textbf{Charge with sign:} \)
\[
q=+3.2\times10^{-13}\,\text{C}
\]
\( \textbf{Interpretation:} \) Missing negative charge leaves a positive net charge.
\( \textbf{Final answer:} \) The net charge is \(+3.2\times10^{-13}\,\text{C}\).
115. The linear charge density \(\lambda\) of a charged wire is defined as charge per unit
ⓐ. mass
ⓑ. area
ⓒ. volume
ⓓ. length
Correct Answer: length
Explanation: Linear charge density is used when charge is distributed along a line-like object such as a thin wire. It is defined by \(\lambda=\frac{dq}{dl}\) for a small length element. For a uniformly charged wire, the simpler form is \(\lambda=\frac{Q}{L}\). Surface charge density uses area, and volume charge density uses volume. Mass is not the geometric measure used in these charge-density definitions. The word linear points to length as the relevant measure.
116. Match each charge density with its defining expression.
| Density | Expression |
|---|
| P. Linear charge density | 1. \(\rho=\frac{dq}{dV}\) |
| Q. Surface charge density | 2. \(\lambda=\frac{dq}{dl}\) |
| R. Volume charge density | 3. \(\sigma=\frac{dq}{dA}\) |
ⓐ. P-1, Q-2, R-3
ⓑ. P-2, Q-3, R-1
ⓒ. P-3, Q-1, R-2
ⓓ. P-2, Q-1, R-3
Correct Answer: P-2, Q-3, R-1
Explanation: Linear charge density describes charge per unit length, so it matches \(\lambda=\frac{dq}{dl}\). Surface charge density describes charge per unit area, so it matches \(\sigma=\frac{dq}{dA}\). Volume charge density describes charge per unit volume, so it matches \(\rho=\frac{dq}{dV}\). The denominators \(dl\), \(dA\), and \(dV\) identify the geometry of the charge distribution. These density symbols are not interchangeable. Choosing the correct density depends on whether the charge is spread along a line, over a surface, or throughout a volume.
117. The unit of surface charge density \(\sigma\) is
ⓐ. \(\text{C m}^{-1}\)
ⓑ. \(\text{C m}^{-3}\)
ⓒ. \(\text{C m}^{-2}\)
ⓓ. \(\text{N C}^{-1}\)
Correct Answer: \(\text{C m}^{-2}\)
Explanation: Surface charge density is charge per unit area. Its defining relation is \(\sigma=\frac{dq}{dA}\). Charge has unit \(\text{C}\), and area has unit \(\text{m}^2\). Therefore, the unit is \(\frac{\text{C}}{\text{m}^2}=\text{C m}^{-2}\). The unit \(\text{C m}^{-1}\) belongs to linear charge density, while \(\text{C m}^{-3}\) belongs to volume charge density. The unit \(\text{N C}^{-1}\) belongs to electric field, not charge density.
118. A uniformly charged thin wire of length \(0.50\,\text{m}\) carries total charge \(2.0\,\mu\text{C}\). Its linear charge density is
ⓐ. \(1.0\,\mu\text{C m}^{-1}\)
ⓑ. \(2.0\,\mu\text{C m}^{-1}\)
ⓒ. \(0.25\,\mu\text{C m}^{-1}\)
ⓓ. \(4.0\,\mu\text{C m}^{-1}\)
Correct Answer: \(4.0\,\mu\text{C m}^{-1}\)
Explanation: \( \textbf{Given charge:} \) \(Q=2.0\,\mu\text{C}\).
\( \textbf{Length of wire:} \) \(L=0.50\,\text{m}\).
\( \textbf{Uniform distribution:} \) Linear charge density can be found using \(\lambda=\frac{Q}{L}\).
\( \textbf{Substitution:} \)
\[
\lambda=\frac{2.0\,\mu\text{C}}{0.50\,\text{m}}
\]
\( \textbf{Calculation:} \)
\[
\lambda=4.0\,\mu\text{C m}^{-1}
\]
\( \textbf{Unit check:} \) Charge divided by length gives \(\text{C m}^{-1}\).
\( \textbf{Why \(1.0\,\mu\text{C m}^{-1}\) fails:} \) It comes from multiplying \(2.0\) by \(0.50\), not dividing by length.
\( \textbf{Final answer:} \) The linear charge density is \(4.0\,\mu\text{C m}^{-1}\).
119. A flat insulating sheet carries charge uniformly over its surface. The most suitable density symbol for describing this distribution is
ⓐ. \(\lambda\)
ⓑ. \(\rho\)
ⓒ. \(\sigma\)
ⓓ. \(k\)
Correct Answer: \(\sigma\)
Explanation: Charge spread over a surface is described by surface charge density. The usual symbol for surface charge density is \(\sigma\). Linear charge density \(\lambda\) is used for line-like charge distributions, such as a thin wire. Volume charge density \(\rho\) is used when charge is distributed throughout a three-dimensional region. The symbol \(k\) is used for Coulomb’s constant, not for charge density. A sheet is a surface distribution, so \(\sigma\) is the natural choice.
120. A non-uniformly charged rod is divided mentally into very small pieces. The use of \(dq\) in \(\lambda=\frac{dq}{dl}\) is helpful because
ⓐ. small lengths may carry different charges
ⓑ. the rod must have zero total charge
ⓒ. charge density is always independent of position
ⓓ. \(dq\) means the rod has no charge
Correct Answer: small lengths may carry different charges
Explanation: For a non-uniform distribution, the charge per unit length may vary from point to point. A small element \(dl\) may carry a small charge \(dq\), so \(\lambda=\frac{dq}{dl}\) gives the local linear charge density. This local idea is more flexible than using only \(\frac{Q}{L}\), which works directly for uniform distributions. A non-uniform rod need not have zero total charge. The symbol \(dq\) means a small amount of charge, not absence of charge. The element method prepares the way for adding contributions from continuous charge distributions.
