201. In the formula \(\vec{p}=q(2\vec{a})\), what does \(2a\) represent?
ⓐ. Distance from the centre of dipole to one charge
ⓑ. Total separation between \(-q\) and \(+q\)
ⓒ. Twice the magnitude of charge
ⓓ. Area enclosed by the dipole
Correct Answer: Total separation between \(-q\) and \(+q\)
Explanation: In the usual notation for an electric dipole, the charges \(-q\) and \(+q\) are separated by a distance \(2a\). The midpoint of the dipole is at a distance \(a\) from each charge. Therefore, \(2a\) is the full dipole length, not the half-length. The magnitude of dipole moment is \(p=q(2a)\). If a student uses \(a\) instead of \(2a\), the calculated dipole moment becomes half of the correct value. The notation mistake is that \(a\) is half-separation, while \(2a\) is the actual charge separation.
202. A dipole has charge magnitude \(q=5.0\times10^{-6}\,\text{C}\) and half-separation \(a=2.0\,\text{cm}\). What is its dipole moment magnitude?
ⓐ. \(1.0\times10^{-7}\,\text{C m}\)
ⓑ. \(2.0\times10^{-7}\,\text{C m}\)
ⓒ. \(5.0\times10^{-8}\,\text{C m}\)
ⓓ. \(2.5\times10^{-4}\,\text{C m}\)
Correct Answer: \(2.0\times10^{-7}\,\text{C m}\)
Explanation: \( \textbf{Given:} \) \(q=5.0\times10^{-6}\,\text{C}\), and \(a=2.0\,\text{cm}=2.0\times10^{-2}\,\text{m}\).
\( \textbf{Important notation check:} \) The separation between the two charges is \(2a\), not \(a\).
\( \textbf{Full separation:} \)
\[
2a=2(2.0\times10^{-2}\,\text{m})=4.0\times10^{-2}\,\text{m}
\]
\( \textbf{Formula:} \)
\[
p=q(2a)
\]
\( \textbf{Substitution:} \)
\[
p=(5.0\times10^{-6})(4.0\times10^{-2})
\]
\( \textbf{Coefficient multiplication:} \)
\[
5.0\times4.0=20.0
\]
\( \textbf{Power multiplication:} \)
\[
10^{-6}\times10^{-2}=10^{-8}
\]
\( \textbf{Simplification:} \)
\[
p=20.0\times10^{-8}\,\text{C m}=2.0\times10^{-7}\,\text{C m}
\]
\( \textbf{common mistake:} \) Using \(a\) directly would give \(1.0\times10^{-7}\,\text{C m}\), which is half the correct value.
\( \textbf{Final answer:} \) The dipole moment magnitude is \(2.0\times10^{-7}\,\text{C m}\).
203. Which option correctly gives the SI unit of electric dipole moment?
ⓐ. \(\text{N m}^2\text{C}^{-1}\)
ⓑ. \(\text{C m}^{-1}\)
ⓒ. \(\text{C m}\)
ⓓ. \(\text{N C}^{-1}\)
Correct Answer: \(\text{C m}\)
Explanation: Electric dipole moment is defined as charge magnitude multiplied by separation between the charges. Its magnitude is \(p=q(2a)\). The SI unit of charge is \(\text{C}\), and the SI unit of length is \(\text{m}\). Therefore, the SI unit of dipole moment is \(\text{C m}\). The unit \(\text{N C}^{-1}\) belongs to electric field, while \(\text{C m}^{-1}\) belongs to linear charge density. A unit check helps separate dipole moment from electric field and charge density.
204. Use the arrangement described below.
On the \(x\)-axis, a charge \(-q\) is fixed at \(x=-a\), and a charge \(+q\) is fixed at \(x=+a\).
What is the direction of the electric dipole moment?
ⓐ. Along \(+x\)-direction
ⓑ. Along \(-x\)-direction
ⓒ. Along \(+y\)-direction
ⓓ. Zero because the net charge is zero
Correct Answer: Along \(+x\)-direction
Explanation: The dipole moment direction is defined from the negative charge toward the positive charge. Here \(-q\) is at \(x=-a\), and \(+q\) is at \(x=+a\). The direction from \(x=-a\) to \(x=+a\) is the positive \(x\)-direction. The net charge of the dipole is zero, but that does not make the dipole moment zero. The separation of equal and opposite charges creates a vector dipole moment. The sign-direction-related mistake is to confuse zero net charge with zero vector moment.
205. Which statement about electric dipole moment is correct?
ⓐ. It is always directed from \(+q\) to \(-q\)
ⓑ. It has unit \(\text{N C}^{-1}\)
ⓒ. It is a scalar quantity because charge is scalar
ⓓ. It is a vector quantity directed from \(-q\) to \(+q\)
Correct Answer: It is a vector quantity directed from \(-q\) to \(+q\)
Explanation: Electric dipole moment is a vector quantity even though charge itself is scalar. The vector nature comes from the displacement from \(-q\) to \(+q\). Its magnitude is \(p=q(2a)\), and its direction is from the negative charge to the positive charge. The unit is \(\text{C m}\), not \(\text{N C}^{-1}\). This vector is important because it determines torque and potential energy of a dipole in an electric field. The common mistake is to reverse the direction or to call \(\vec{p}\) scalar because \(q\) is scalar.
206. Study the table and identify the row with the correct description.
| Row | Dipole quantity | Description |
|---|
| P | Net charge of dipole | \(+2q\) |
| Q | Dipole moment magnitude | \(p=q(2a)\) |
| R | Direction of \(\vec{p}\) | From \(+q\) to \(-q\) |
| S | Unit of \(\vec{p}\) | \(\text{N C}^{-1}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row Q
Explanation: Row Q is correct because the magnitude of electric dipole moment is \(p=q(2a)\). Row P is incorrect because the net charge of a dipole is \(+q+(-q)=0\), not \(+2q\). Row R is incorrect because \(\vec{p}\) is directed from \(-q\) to \(+q\). Row S is incorrect because the unit of dipole moment is \(\text{C m}\), while \(\text{N C}^{-1}\) is the unit of electric field. The table mainly tests whether charge separation, direction convention, and units are kept distinct. The safest memory is \(p=\text{charge}\times\text{separation}\), directed from negative to positive.
207. Consider the following statements about an electric dipole.
Statement I: Its two charges are equal in magnitude and opposite in sign.
Statement II: Its net charge is zero.
Statement III: Its dipole moment must be zero because its net charge is zero.
Statement IV: Its dipole moment has direction from \(-q\) to \(+q\).
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is correct because a dipole is made of equal and opposite charges. Statement II is correct because the algebraic sum of \(+q\) and \(-q\) is zero. Statement III is incorrect because zero net charge does not imply zero dipole moment when the charges are separated. Statement IV is correct because the conventional direction of \(\vec{p}\) is from the negative charge to the positive charge. The separation between the charges is essential in deciding the dipole moment. The main misconception is to judge dipole moment only from net charge and ignore charge separation.
208. A pair of charges \(+q\) and \(-q\) is separated by distance \(d\). If \(q\) is doubled and \(d\) is halved, what happens to the dipole moment magnitude?
ⓐ. It becomes twice
ⓑ. It becomes half
ⓒ. It remains unchanged
ⓓ. It becomes four times
Correct Answer: It remains unchanged
Explanation: \( \textbf{Initial dipole moment:} \)
\[
p=qd
\]
where \(d\) is the separation between the charges.
\( \textbf{Changed charge:} \) The charge magnitude becomes \(2q\).
\( \textbf{Changed separation:} \) The separation becomes \(\frac{d}{2}\).
\( \textbf{New dipole moment:} \)
\[
p'=(2q)\left(\frac{d}{2}\right)
\]
\( \textbf{Simplification:} \)
\[
p'=qd
\]
\( \textbf{Comparison:} \)
\[
p'=p
\]
\( \textbf{Physical meaning:} \) The increase in charge magnitude is exactly balanced by the decrease in separation.
\( \textbf{Final answer:} \) The dipole moment magnitude remains unchanged.
209. Read the situation below and answer the question.
A neutral molecule has centres of positive and negative charge that do not coincide. The positive and negative charge magnitudes are equal, but their centres are separated by a small distance.
Which statement best describes the molecule in electrostatic terms?
ⓐ. It has no electric effect because its net charge is zero
ⓑ. It behaves like an electric dipole
ⓒ. It must be a positive point charge
ⓓ. It must have zero dipole moment because both charge magnitudes are equal
Correct Answer: It behaves like an electric dipole
Explanation: A system can have zero net charge and still behave as an electric dipole if equal positive and negative charge centres are separated. The molecule described has equal and opposite charge centres, so its net charge is zero. However, the centres do not coincide, so it has a non-zero dipole moment. This is the basic idea behind polar molecules. The electric dipole moment depends on both charge magnitude and separation. A common mistake is assuming that neutral automatically means no dipole behaviour.
210. What is meant by the ideal dipole approximation?
ⓐ. The two charges become unequal while separation becomes very large
ⓑ. The charge separation is small compared with the observation distance
ⓒ. The two charges are placed at the same point and the dipole moment becomes maximum
ⓓ. The net charge becomes \(+q\) instead of zero
Correct Answer: The charge separation is small compared with the observation distance
Explanation: An ideal dipole is a useful approximation for studying fields at distances much larger than the charge separation. The charges are equal and opposite, and their separation is considered very small compared with the observation distance. The dipole moment \(p=q(2a)\) is treated as finite in the approximation. This allows simpler far-field expressions for the electric field of a dipole. The approximation does not mean the two charges are unequal or that the net charge becomes non-zero. The condition to remember is small separation compared with the distance of the field point.
211. A dipole has moment \(\vec{p}=8.0\times10^{-8}\hat{i}\,\text{C m}\). If its charge magnitude is \(4.0\,\mu\text{C}\), what is the separation between its charges?
ⓐ. \(2.0\times10^{-1}\,\text{m}\)
ⓑ. \(5.0\times10^{1}\,\text{m}\)
ⓒ. \(2.0\times10^{-2}\,\text{m}\)
ⓓ. \(3.2\times10^{-13}\,\text{m}\)
Correct Answer: \(2.0\times10^{-2}\,\text{m}\)
Explanation: \( \textbf{Given:} \) Dipole moment magnitude \(p=8.0\times10^{-8}\,\text{C m}\), and charge magnitude \(q=4.0\,\mu\text{C}=4.0\times10^{-6}\,\text{C}\).
\( \textbf{Required:} \) Separation \(d\) between the charges.
\( \textbf{Dipole moment relation:} \)
\[
p=qd
\]
where \(d=2a\) is the full separation.
\( \textbf{Rearranged form:} \)
\[
d=\frac{p}{q}
\]
\( \textbf{Substitution:} \)
\[
d=\frac{8.0\times10^{-8}}{4.0\times10^{-6}}
\]
\( \textbf{Coefficient simplification:} \)
\[
\frac{8.0}{4.0}=2.0
\]
\( \textbf{Power simplification:} \)
\[
\frac{10^{-8}}{10^{-6}}=10^{-2}
\]
\( \textbf{Result:} \)
\[
d=2.0\times10^{-2}\,\text{m}
\]
\( \textbf{Direction note:} \) The \(\hat{i}\) direction tells the direction from \(-q\) to \(+q\), not the separation magnitude.
\( \textbf{Final answer:} \) The charge separation is \(2.0\times10^{-2}\,\text{m}\).
212. Which statement correctly compares an electric dipole with a single point charge at large distances?
ⓐ. A dipole moment is meaningful only when both charges have the same sign
ⓑ. A dipole has non-zero net charge, so it always behaves exactly like a point charge
ⓒ. A dipole has no electric field anywhere because its charges cancel
ⓓ. A dipole has zero net charge but non-zero dipole moment, so its far field differs
Correct Answer: A dipole has zero net charge but non-zero dipole moment, so its far field differs
Explanation: A single point charge has non-zero net charge and produces a field varying as \(r^{-2}\). An electric dipole has zero net charge because its charges are equal and opposite. However, the charges are separated, so the dipole has a non-zero dipole moment. At large distances, the dipole field decreases faster than the field of a single point charge. This difference arises because the leading net-charge effect cancels, leaving the dipole effect. The misconception to avoid is that zero net charge means zero field at all points.
213. On the axial line of an electric dipole, the far-field magnitude is approximately:
ⓐ. \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}\)
ⓑ. \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\)
ⓒ. \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{p}{r}\)
ⓓ. \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^2}\)
Correct Answer: \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\)
Explanation: The axial line of a dipole is the line passing through both charges of the dipole. For a point far from the dipole, where \(r\) is much greater than the dipole half-separation \(a\), the axial field has the approximate magnitude \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\). This shows that the far field of a dipole varies as \(r^{-3}\), not \(r^{-2}\). The factor \(2\) is specific to the axial line. A point charge field varies as \(r^{-2}\), so using \(r^2\) here is a common confusion. The far-field condition is important because the expression is an approximation for \(r\gg a\).
214. For a short electric dipole, the far axial electric field varies with distance \(r\) as:
ⓐ. \(E\propto r^{-1}\)
ⓑ. \(E\propto r^{-2}\)
ⓒ. \(E\propto r^{-3}\)
ⓓ. \(E\propto r^{2}\)
Correct Answer: \(E\propto r^{-3}\)
Explanation: On the far axial line of an electric dipole, the approximate field is \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\). For a fixed dipole moment \(p\), the only varying factor is \(\frac{1}{r^3}\). Therefore, the far axial field follows \(E\propto r^{-3}\). This is a faster decrease than the field of a point charge, which follows \(E\propto r^{-2}\). The result is valid in the far-field approximation where \(r\) is much larger than the dipole separation. The common mistake is to apply the point-charge inverse-square law directly to a dipole.
215. Which expression gives the exact electric field magnitude on the axial line of a dipole at a point distance \(r\) from its centre, where the dipole charges are separated by \(2a\) and \(r\gt a\)?
ⓐ. \(E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}}\)
ⓑ. \(E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{2pr}{(r^2-a^2)^2}\)
ⓒ. \(E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}\)
ⓓ. \(E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r}\)
Correct Answer: \(E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{2pr}{(r^2-a^2)^2}\)
Explanation: For a dipole with charges \(+q\) and \(-q\) separated by \(2a\), the dipole moment is \(p=q(2a)\). At an axial point distance \(r\) from the centre, the distances from the two charges are \(r-a\) and \(r+a\). The field contributions are along the same line but in opposite directions, so their magnitudes are subtracted carefully. The exact simplification gives \(E_{\text{axial}}=\frac{1}{4\pi\varepsilon_0}\frac{2pr}{(r^2-a^2)^2}\). This expression is valid for an axial point outside the dipole, where \(r\gt a\). The far-field form \(\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\) is obtained only when \(r\gg a\).
216. Which condition is required to use the far-field approximation \(E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\) for a dipole?
ⓐ. \(r\ll a\)
ⓑ. \(r=a\)
ⓒ. \(r\gg a\)
ⓓ. \(r=0\)
Correct Answer: \(r\gg a\)
Explanation: The exact axial field of a dipole contains terms involving both \(r\) and \(a\). The far-field approximation is made when the observation point is very far from the dipole compared with the half-separation \(a\). In that case, \(r^2-a^2\) is approximately \(r^2\), so the exact expression reduces to a simpler \(r^{-3}\) form. This is why the condition is \(r\gg a\). If \(r\) is comparable to \(a\), the exact expression should be used. The approximation mistake is to apply the far-field formula near the dipole charges.
217. A dipole has moment \(p=4.0\times10^{-8}\,\text{C m}\). What is the approximate electric field magnitude at an axial point \(0.20\,\text{m}\) from its centre, assuming \(r\gg a\)? Take \(\frac{1}{4\pi\varepsilon_0}=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
ⓐ. \(4.5\times10^4\,\text{N C}^{-1}\)
ⓑ. \(9.0\times10^4\,\text{N C}^{-1}\)
ⓒ. \(1.8\times10^5\,\text{N C}^{-1}\)
ⓓ. \(3.6\times10^5\,\text{N C}^{-1}\)
Correct Answer: \(9.0\times10^4\,\text{N C}^{-1}\)
Explanation: \( \textbf{Given:} \) \(p=4.0\times10^{-8}\,\text{C m}\), \(r=0.20\,\text{m}\), and \(k=\frac{1}{4\pi\varepsilon_0}=9.0\times10^9\,\text{N m}^2\text{C}^{-2}\).
\( \textbf{Required:} \) Far axial electric field magnitude.
\( \textbf{Formula:} \)
\[
E_{\text{axial}}\approx k\frac{2p}{r^3}
\]
\( \textbf{Why it applies:} \) The question states \(r\gg a\), so the far-field axial approximation is valid.
\( \textbf{Cube of distance:} \)
\[
r^3=(0.20)^3=0.008\,\text{m}^3
\]
\( \textbf{Numerator:} \)
\[
2p=2(4.0\times10^{-8})=8.0\times10^{-8}\,\text{C m}
\]
\( \textbf{Substitution:} \)
\[
E_{\text{axial}}=\frac{(9.0\times10^9)(8.0\times10^{-8})}{0.008}
\]
\( \textbf{Numerator simplification:} \)
\[
(9.0\times10^9)(8.0\times10^{-8})=72.0\times10^1=720
\]
\( \textbf{Final calculation:} \)
\[
E_{\text{axial}}=\frac{720}{0.008}=9.0\times10^4\,\text{N C}^{-1}
\]
\( \textbf{Final answer:} \) The approximate axial field magnitude is \(9.0\times10^4\,\text{N C}^{-1}\).
218. If the distance from the centre of a short dipole is doubled on the far axial line, how does the electric field magnitude change?
ⓐ. It becomes \(\frac{1}{2}\) of the original
ⓑ. It becomes \(\frac{1}{4}\) of the original
ⓒ. It becomes \(\frac{1}{8}\) of the original
ⓓ. It becomes \(8\) times the original
Correct Answer: It becomes \(\frac{1}{8}\) of the original
Explanation: On the far axial line of a short dipole, \(E_{\text{axial}}\propto \frac{1}{r^3}\). If the distance becomes \(2r\), the new field is proportional to \(\frac{1}{(2r)^3}\). Since \((2r)^3=8r^3\), the field becomes \(\frac{1}{8}\) of its original value. This is different from the point-charge field, which would become \(\frac{1}{4}\) on doubling distance. The faster fall is a characteristic feature of dipole fields. A common mistake is to use inverse-square dependence instead of inverse-cube dependence.
219. A graph is plotted between far axial electric field \(E_{\text{axial}}\) of a fixed dipole and \(\frac{1}{r^3}\). What does the slope represent?
ⓐ. \(\frac{p}{4\pi\varepsilon_0}\)
ⓑ. \(\frac{r^3}{4\pi\varepsilon_0p}\)
ⓒ. \(\frac{p}{2r^3}\)
ⓓ. \(\frac{2p}{4\pi\varepsilon_0}\)
Correct Answer: \(\frac{2p}{4\pi\varepsilon_0}\)
Explanation: \( \textbf{Far axial field:} \)
\[
E_{\text{axial}}\approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}
\]
\( \textbf{Graph variable:} \) The horizontal axis is \(\frac{1}{r^3}\).
\( \textbf{Straight-line form:} \)
\[
E_{\text{axial}}=\left(\frac{2p}{4\pi\varepsilon_0}\right)\left(\frac{1}{r^3}\right)
\]
\( \textbf{Comparison with} \) \(y=mx\): \(E_{\text{axial}}\) behaves like \(y\), and \(\frac{1}{r^3}\) behaves like \(x\).
\( \textbf{Slope identification:} \) The coefficient of \(\frac{1}{r^3}\) is \(\frac{2p}{4\pi\varepsilon_0}\).
\( \textbf{Condition check:} \) This interpretation is valid only for the far axial region of a fixed dipole.
\( \textbf{Final answer:} \) The slope represents \(\frac{2p}{4\pi\varepsilon_0}\).
220. On the axial line of a dipole, why is the far-field dependence \(E\propto r^{-3}\) instead of the point-charge dependence \(E\propto r^{-2}\)?
ⓐ. Because the two charges in a dipole have the same sign
ⓑ. Because the dipole moment is always zero
ⓒ. Because electric field of every charge distribution varies as \(r^{-3}\)
ⓓ. Because opposite charges cancel the leading \(r^{-2}\) term
Correct Answer: Because opposite charges cancel the leading \(r^{-2}\) term
Explanation: A single point charge has non-zero net charge, so its field decreases as \(r^{-2}\). A dipole has equal and opposite charges, so its net charge is zero. At large distances, the leading contribution corresponding to total charge cancels. The remaining effect depends on charge separation, represented by the dipole moment \(\vec{p}\). This remaining dipole field falls faster, as \(r^{-3}\). The misconception is to think the two charges cancel the field everywhere; they cancel the net-charge term, not the entire electric field.
