Testing 11 Types of MCQs

Explanation: Electric charge is treated as a scalar physical quantity in elementary electrostatics. The signs \(+\) and \(-\) distinguish two kinds of charge, but they do not represent directions in space. A vector quantity needs a spatial direction, such as along an axis or along a field, while charge has no such direction. A charge such as \(+3\,\text{C}\) and a charge such as \(-3\,\text{C}\) differ in type, not in vector direction. The sign is important because it helps decide whether two charges attract or repel. Therefore, electric charge is a scalar quantity with positive or negative sign.

Explanation: Electric field lines never intersect in electrostatics. The tangent to a field line at any point gives the direction of \(\vec{E}\) at that point. If two field lines intersected, two different tangents could be drawn at the same point. That would mean two different directions of electric field at one point, which is not physically possible for a unique electric field. The Reason gives the exact explanation of why the Assertion is true. Hence, both Assertion and Reason are true, and the Reason explains the Assertion.

Explanation: \( \textbf{Given:} \) \(q_1=2\times10^{-6}\,\text{C}\), \(q_2=6\times10^{-6}\,\text{C}\), \(r=0.30\,\text{m}\), \(K=3\), \(m=2.0\times10^{-4}\,\text{kg}\). \( \textbf{Required:} \) Acceleration of the body carrying \(+2\,\mu\text{C}\). \( \textbf{Force in medium:} \) \[ F=\frac{k|q_1q_2|}{Kr^2} \] \( \textbf{Substitution:} \) \[ F=\frac{(9.0\times10^9)(2\times10^{-6})(6\times10^{-6})}{3(0.30)^2} \] \( \textbf{Numerator:} \) \[ (9.0\times10^9)(12\times10^{-12})=0.108 \] \( \textbf{Denominator:} \) \[ 3(0.30)^2=3(0.09)=0.27 \] \( \textbf{Electrostatic force:} \) \[ F=\frac{0.108}{0.27}=0.40\,\text{N} \] \( \textbf{Acceleration relation:} \) \[ a=\frac{F}{m} \] \( \textbf{Final calculation:} \) \[ a=\frac{0.40}{2.0\times10^{-4}}=2.0\times10^3\,\text{m s}^{-2} \] \( \textbf{Final answer:} \) The acceleration is \(2.0\times10^3\,\text{m s}^{-2}\).

Explanation: Charge quantisation means that charge occurs in integral multiples of elementary charge, so it matches \(q=ne\), where \(n\in\mathbb{Z}\). Conservation of charge means that the net charge of an isolated system remains constant. Additivity of charge means that the total charge is the algebraic sum \(Q=q_1+q_2+q_3+\cdots\). In a dielectric medium of dielectric constant \(K\), Coulomb force becomes \(F_m=\frac{F_{\text{vacuum}}}{K}\). Thus, the correct matching is P-3, Q-4, R-2, S-1.

Explanation: Electric field at a point is defined as force per unit positive test charge, \( \vec{E}=\frac{\vec{F}}{q_0} \). The test charge is only a probe used to detect the field produced by the source charges. If the test charge is large, it can exert a significant force on the source charges and disturb their arrangement. A disturbed source distribution would change the field being measured, so the measurement would no longer represent the original field. The test charge is taken positive so that the direction of electric field matches the direction of force on it. Therefore, the test charge should be very small and positive.

Explanation: For a point charge, electric field varies as \(E\propto r^{-2}\), so row P is correct. Far from an electric dipole, the field varies as \(E\propto r^{-3}\), so row Q is also correct. For an infinitely long line charge, \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\), so \(E\propto r^{-1}\), making row R correct. A single infinite sheet has field \(E=\frac{\sigma}{2\varepsilon_0}\), which is independent of distance from the sheet. Therefore, row S is incorrect because it wrongly assigns inverse-square variation to a single infinite sheet.

Explanation: Quantisation of charge means that any isolated charge can be written as \(q=ne\), where \(n\) is an integer and \(e\) is the elementary charge. The elementary charge has magnitude \(1.6\times10^{-19}\,\text{C}\). This rule does not involve \(\varepsilon_0\), because \(\varepsilon_0\) is permittivity. It also does not involve \(k\), which is Coulomb's constant, or \(\lambda\), which is linear charge density. Therefore, the blank must be \(e\).

Explanation: Electric field due to a positive charge points away from the positive charge, so at the midpoint \(M\), the field due to \(+q\) points from \(+q\) toward \(-q\). Electric field due to a negative charge points toward the negative charge, so at \(M\), the field due to \(-q\) also points toward \(-q\). Thus, the two field contributions are in the same direction and add. Equal magnitudes of the charges do not imply zero field here because the two fields are not opposite at the midpoint. The net electric field at \(M\) is from \(+q\) toward \(-q\).

Explanation: For two fixed point charges in vacuum, Coulomb's law gives \(F=k\frac{|q_1q_2|}{r^2}\). Since the horizontal axis is \(\frac{1}{r^2}\), the equation can be read as \(F=(k|q_1q_2|)\left(\frac{1}{r^2}\right)\). This has the form \(y=mx\), where \(F\) behaves like \(y\), \(\frac{1}{r^2}\) behaves like \(x\), and \(k|q_1q_2|\) is the slope. Since the charges are fixed, this coefficient remains constant. Therefore, the slope of the graph represents \(k|q_1q_2|\).

Explanation: Electric flux is a scalar quantity because it is obtained from the dot product \(\vec{E}\cdot\vec{A}\). In the formula \(\Phi_E=EA\cos\theta\), the angle \(\theta\) is measured between the electric field vector and the area vector, not between the field and the surface plane. For a closed surface, the area vector is taken outward by convention. Therefore, field lines leaving the surface contribute positive flux, while field lines entering the surface contribute negative flux. All three statements are correct.

Explanation: The centre \(O\) is equidistant from all four corners of the square. The top-left \(+q\) and bottom-right \(+q\) produce equal electric fields at \(O\) in opposite directions along one diagonal, so their fields cancel. The top-right \(-q\) and bottom-left \(-q\) also produce equal fields at \(O\) in opposite directions along the other diagonal, so they also cancel. Electric field is a vector quantity, so these directional cancellations are important. Therefore, the net electric field at \(O\) is zero.

Explanation: \( \textbf{Given:} \) \(q=3\times10^{-6}\,\text{C}\), separation \(=8\,\text{cm}=0.08\,\text{m}\), \(E=2.0\times10^5\,\text{N C}^{-1}\), \(\theta=30^\circ\), and \(I=4.8\times10^{-5}\,\text{kg m}^2\). \( \textbf{Required:} \) Initial angular acceleration \(\alpha\). \( \textbf{Dipole moment:} \) \[ p=q(2a) \] Here the separation between charges is already \(2a=0.08\,\text{m}\). \( \textbf{Calculation of \(p\):} \) \[ p=(3\times10^{-6})(0.08)=2.4\times10^{-7}\,\text{C m} \] \( \textbf{Torque on dipole:} \) \[ \tau=pE\sin\theta \] \( \textbf{Substitution:} \) \[ \tau=(2.4\times10^{-7})(2.0\times10^5)\sin30^\circ \] \( \textbf{Torque value:} \) \[ \tau=(2.4\times10^{-7})(2.0\times10^5)\left(\frac{1}{2}\right)=2.4\times10^{-2}\,\text{N m} \] \( \textbf{Rotational relation:} \) \[ \tau=I\alpha \] \( \textbf{Angular acceleration:} \) \[ \alpha=\frac{\tau}{I}=\frac{2.4\times10^{-2}}{4.8\times10^{-5}}=5.0\times10^2\,\text{rad s}^{-2} \] \( \textbf{Final answer:} \) The initial angular acceleration is \(5.0\times10^2\,\text{rad s}^{-2}\).

Explanation: Electric field direction is defined using a small positive test charge. If a positive test charge is placed at a point, the direction of force on it gives the direction of \(\vec{E}\) at that point. A negative charge would experience force opposite to the electric field direction, so it is not used for defining the field direction. A neutral body cannot be used in the definition because the electric force on it is not represented by \(q_0\vec{E}\). Therefore, the blank is positive.

Explanation: Charging by induction occurs without direct contact between the charged body and the conductor being charged. When a charged rod is brought near a neutral conductor, charges inside the conductor redistribute. Earthing then allows electrons to enter or leave the conductor. If the earth connection is removed before the inducing rod is taken away, the conductor can be left with a permanent charge. Rubbing glass with silk is charging by friction, and touching a charged sphere to a neutral sphere is charging by conduction. Therefore, the described earthing sequence represents charging by induction.

Explanation: The symbol \(\lambda\) represents linear charge density, so P matches 3. The symbol \(\sigma\) represents surface charge density, so Q matches 4. The symbol \(\rho\) represents volume charge density, so R matches 2. The symbol \(\Phi_E\) is used for electric flux, so S matches 1. The unit check also supports this: \(\lambda\) has unit \(\text{C m}^{-1}\), \(\sigma\) has unit \(\text{C m}^{-2}\), and \(\rho\) has unit \(\text{C m}^{-3}\). Hence, the matching is P-3, Q-4, R-2, S-1.

Explanation: Gauss's law states that the net electric flux through a closed surface is \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\). Only the charge enclosed by the closed surface contributes to the net flux. The charge \(+Q\) is inside \(S\), while \(+2Q\) is outside it. The outside charge can change the electric field at different points on the surface, but its net contribution to the closed-surface flux is zero. Therefore, the net electric flux through \(S\) is \(\frac{Q}{\varepsilon_0}\).

Explanation: A dipole consists of equal and opposite charges separated by a small distance. In a uniform electric field, \(+q\) experiences force along the field and \(-q\) experiences force opposite to the field. These two forces have equal magnitude \(qE\) and opposite directions, so their vector sum is zero. The dipole can still experience torque if it is not aligned with the field, but its net force in a uniform field is zero. The Reason correctly explains the Assertion. Hence, both Assertion and Reason are true, and the Reason explains the Assertion.

Explanation: For an infinitely long uniformly charged wire, the electric field at distance \(r\) is \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\). Since the graph is drawn between \(E\) and \(\frac{1}{r}\), the relation can be written as \(E=\left(\frac{\lambda}{2\pi\varepsilon_0}\right)\left(\frac{1}{r}\right)\). Comparing this with \(y=mx\), the slope is \(\frac{\lambda}{2\pi\varepsilon_0}\). This graph is different from the point-charge field graph because a line charge gives \(E\propto \frac{1}{r}\), not \(E\propto \frac{1}{r^2}\). Therefore, the slope represents \(\frac{\lambda}{2\pi\varepsilon_0}\).

Explanation: According to Gauss's law, the net flux through a closed surface is \(\Phi_E=\frac{q_{\text{enc}}}{\varepsilon_0}\). Surface S1 encloses \(+q\) and \(-q\), so its net enclosed charge is zero. Therefore, the net flux through S1 is also zero. Surface S2 encloses \(+2q\), surface S3 encloses \(-q\), and surface S4 encloses \(+2q\), so none of them has zero net enclosed charge. Hence, S1 has zero net electric flux.

Explanation: Gauss's law is valid for every closed surface, but it becomes directly useful for finding electric field only when there is enough symmetry. The law is \(\oint\vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{\varepsilon_0}\). To obtain \(E\) easily, the field magnitude must be constant over useful parts of the Gaussian surface or the flux through some parts must be zero by symmetry. Without such symmetry, \(E\) changes from point to point and cannot be taken out of the integral. External charges do not invalidate Gauss's law, but they can make the local field more complicated. Therefore, suitable symmetry is needed to use Gauss's law as a practical field-finding tool.