Class 12 Physics MCQs | Chapter 1: Electric Charges and Fields – Part 2

Explanation: \( \textbf{Given:} \) Two force magnitudes are \(6\,\text{N}\) and \(8\,\text{N}\), and they are perpendicular. \( \textbf{Required:} \) Magnitude of the resultant force. \( \textbf{Vector addition reason:} \) Forces are vectors, so perpendicular forces are not added by simple arithmetic. \( \textbf{Resultant formula:} \) \[ F_{\text{net}}=\sqrt{F_1^2+F_2^2} \] \( \textbf{Substitution:} \) \[ F_{\text{net}}=\sqrt{(6)^2+(8)^2} \] \( \textbf{Intermediate simplification:} \) \[ F_{\text{net}}=\sqrt{36+64} \] \( \textbf{Final simplification:} \) \[ F_{\text{net}}=\sqrt{100}=10\,\text{N} \] \( \textbf{Direction note:} \) The resultant lies between the two perpendicular force directions. \( \textbf{Final answer:} \) The net force magnitude is \(10\,\text{N}\).

Explanation: \( \textbf{Known force components:} \) The force along \(+x\) is \(3\,\text{N}\), and the force along \(+y\) is \(4\,\text{N}\). \( \textbf{Vector nature:} \) Since the forces are perpendicular, their resultant is found using the Pythagorean relation. \( \textbf{Magnitude formula:} \) \[ F_{\text{net}}=\sqrt{F_x^2+F_y^2} \] \( \textbf{Substitution:} \) \[ F_{\text{net}}=\sqrt{3^2+4^2} \] \( \textbf{Simplification:} \) \[ F_{\text{net}}=\sqrt{9+16}=\sqrt{25}=5\,\text{N} \] \( \textbf{Direction check:} \) Both components are positive, so the resultant points in the first quadrant. \( \textbf{common mistake:} \) Adding \(3\,\text{N}+4\,\text{N}=7\,\text{N}\) would be valid only if both forces were collinear in the same direction. \( \textbf{Final answer:} \) The net force is \(5\,\text{N}\) in the first quadrant.

Explanation: \( \textbf{Given forces:} \) One force is \(12\,\text{N}\) to the right, and the other is \(5\,\text{N}\) to the left. \( \textbf{Choose sign convention:} \) Take right as positive and left as negative. \( \textbf{Write signed forces:} \) \[ F_1=+12\,\text{N},\qquad F_2=-5\,\text{N} \] \( \textbf{Vector sum along a line:} \) \[ F_{\text{net}}=F_1+F_2 \] \( \textbf{Substitution:} \) \[ F_{\text{net}}=+12\,\text{N}-5\,\text{N} \] \( \textbf{Simplification:} \) \[ F_{\text{net}}=+7\,\text{N} \] \( \textbf{Direction check:} \) The positive sign means the resultant is to the right. \( \textbf{Final answer:} \) The net force is \(7\,\text{N}\) to the right.

Explanation: Superposition requires treating each pairwise interaction separately. To find the net force on a chosen charge, first calculate or identify the force due to each other charge. Each force must include both magnitude and direction. After that, the individual forces are added as vectors. Algebraically adding the source charges is generally not valid unless symmetry or a special equivalent-charge argument supports it. A common mistake is to replace a vector addition problem with a simple scalar addition of charges or force magnitudes.

Explanation: \( \textbf{Given:} \) Central charge \(q_0=+2\times10^{-6}\,\text{C}\), right charge \(q_R=+3\times10^{-6}\,\text{C}\), left charge \(q_L=-3\times10^{-6}\,\text{C}\), and \(r=0.30\,\text{m}\). \( \textbf{Required:} \) Net force on the charge at the origin. \( \textbf{Force due to right charge:} \) The right charge is positive, so it repels the central positive charge toward \(-x\). \( \textbf{Force due to left charge:} \) The left charge is negative, so it attracts the central positive charge toward \(-x\). \( \textbf{Magnitude of each force:} \) \[ F=k\frac{|q_0q|}{r^2} \] \( \textbf{Substitution:} \) \[ F=\frac{(9.0\times10^9)(2\times10^{-6})(3\times10^{-6})}{(0.30)^2} \] \( \textbf{Simplification:} \) \[ F=\frac{(9.0\times10^9)(6\times10^{-12})}{0.090}=\frac{0.054}{0.090}=0.60\,\text{N} \] \( \textbf{Vector addition:} \) Both forces are along \(-x\), so they add. \[ F_{\text{net}}=0.60\,\text{N}+0.60\,\text{N}=1.20\,\text{N} \] \( \textbf{Final answer:} \) The net force is \(1.20\,\text{N}\) along \(-x\)-direction.

Explanation: The central charge is equidistant from all four corner charges. Each corner charge repels the central positive charge along the line joining that corner and the centre. The force due to a corner charge is cancelled by the force due to the diagonally opposite corner charge. The same cancellation happens for the other diagonal pair. Since all four charges are identical and symmetrically placed, the vector sum of the four forces is zero. A common mistake is to add four equal magnitudes directly; symmetry requires vector cancellation, not scalar addition.

Explanation: The charge on the positive \(x\)-axis repels the central \(+q\), so its force on \(+q\) is along \(-x\). The charge on the positive \(y\)-axis also repels the central \(+q\), so its force on \(+q\) is along \(-y\). The two source charges have equal magnitude and are at the same distance, so the two force magnitudes are equal. Equal perpendicular forces combine along the angle bisector between their directions. Therefore, the resultant lies midway between \(-x\) and \(-y\). The common mistake is to point the resultant toward the source charges, but repulsion pushes \(+q\) away from both positive charges.

Explanation: \( \textbf{Given:} \) Two forces of equal magnitude \(F\) act perpendicular to each other. \( \textbf{Required:} \) Magnitude of the resultant force. \( \textbf{Vector addition rule:} \) Perpendicular forces are combined by the Pythagorean relation. \[ F_{\text{net}}=\sqrt{F_x^2+F_y^2} \] \( \textbf{Substitution:} \) \[ F_{\text{net}}=\sqrt{F^2+F^2} \] \( \textbf{Simplification:} \) \[ F_{\text{net}}=\sqrt{2F^2} \] \[ F_{\text{net}}=\sqrt{2}F \] \( \textbf{Direction note:} \) Since the forces are equal, the resultant is along the angle bisector of \(+x\) and \(+y\). \( \textbf{Final answer:} \) The net force magnitude is \(\sqrt{2}F\).

Explanation: \( \textbf{Given:} \) \(q_0=1.0\times10^{-6}\,\text{C}\), \(q_x=3.0\times10^{-6}\,\text{C}\), \(r_x=0.30\,\text{m}\), \(q_y=4.0\times10^{-6}\,\text{C}\), and \(r_y=0.40\,\text{m}\). \( \textbf{Required:} \) Magnitude of net force on \(q_0\). \( \textbf{Force due to charge on} \) \(x\)-axis: \[ F_x=k\frac{|q_0q_x|}{r_x^2} \] \( \textbf{Substitution for} \) \(F_x\): \[ F_x=\frac{(9.0\times10^9)(1.0\times10^{-6})(3.0\times10^{-6})}{(0.30)^2} \] \( \textbf{Simplification:} \) \[ F_x=\frac{0.027}{0.090}=0.300\,\text{N} \] \( \textbf{Force due to charge on} \) \(y\)-axis: \[ F_y=k\frac{|q_0q_y|}{r_y^2} \] \( \textbf{Substitution for} \) \(F_y\): \[ F_y=\frac{(9.0\times10^9)(1.0\times10^{-6})(4.0\times10^{-6})}{(0.40)^2} \] \( \textbf{Simplification:} \) \[ F_y=\frac{0.036}{0.160}=0.225\,\text{N} \] \( \textbf{Resultant magnitude:} \) \[ F_{\text{net}}=\sqrt{(0.300)^2+(0.225)^2} \] \[ F_{\text{net}}=\sqrt{0.0900+0.050625}=\sqrt{0.140625}=0.375\,\text{N} \] \( \textbf{Final answer:} \) The magnitude of net force is \(0.375\,\text{N}\).

Explanation: The superposition principle says that the force due to each charge is calculated as if the other charges do not disturb that particular pairwise interaction. Other charges add their own forces, but they do not switch off the force between a selected pair. The net force is found only after all individual force vectors are added. The nearest charge may produce a large force, but farther charges can still contribute. A many-charge system does not automatically have zero net force. The key point is independence of individual interactions followed by vector addition.

Explanation: \( \textbf{Given:} \) Two equal forces of magnitude \(F\) act on \(+q\). \( \textbf{Triangle geometry:} \) In an equilateral triangle, the angle between the two force directions at a vertex is \(60^\circ\). \( \textbf{Resultant formula:} \) \[ R=\sqrt{F^2+F^2+2F^2\cos60^\circ} \] \( \textbf{Use} \) \(\cos60^\circ=\frac{1}{2}\): \[ R=\sqrt{F^2+F^2+2F^2\left(\frac{1}{2}\right)} \] \( \textbf{Simplification:} \) \[ R=\sqrt{3F^2} \] \[ R=\sqrt{3}F \] \( \textbf{Direction meaning:} \) The resultant lies along the angle bisector of the two equal force directions. \( \textbf{Final answer:} \) The net force magnitude is \(\sqrt{3}F\).

Explanation: The negative charge at the origin is attracted toward the positive charge at \(x=-a\), producing a force along \(-x\). It is also attracted toward the positive charge at \(x=+a\), producing an equal force along \(+x\). The two positive charges are equal and placed at equal distances from the origin. Therefore, the two force magnitudes on \(-q\) are equal. Since the directions are opposite, the vector sum is zero. The central charge is still acted on by both charges individually; the net force is zero only because of symmetric cancellation.

Explanation: Row P is correct because two forces in the same direction add directly, giving \(8\,\text{N}\) right. Row Q is wrong because opposite collinear forces subtract, so the resultant should be \(2\,\text{N}\) right. Row R is wrong because perpendicular forces must be added by \(\sqrt{5^2+3^2}\), not by simple scalar addition. Row S is wrong because equal opposite forces cancel, giving zero resultant. The table highlights why force addition must respect direction. The common mistake is to add magnitudes without checking whether the forces are same-direction, opposite-direction, or perpendicular.

Explanation: \( \textbf{Given components:} \) \(F_x=12\,\text{N}\) and \(F_y=5\,\text{N}\). \( \textbf{Vector-diagram meaning:} \) The diagonal represents the resultant of two perpendicular components. \( \textbf{Formula:} \) \[ F_{\text{net}}=\sqrt{F_x^2+F_y^2} \] \( \textbf{Substitution:} \) \[ F_{\text{net}}=\sqrt{(12)^2+(5)^2} \] \( \textbf{Intermediate simplification:} \) \[ F_{\text{net}}=\sqrt{144+25} \] \( \textbf{Final simplification:} \) \[ F_{\text{net}}=\sqrt{169}=13\,\text{N} \] \( \textbf{Graph interpretation:} \) The diagonal is longer than either component but shorter than their direct scalar sum \(17\,\text{N}\). \( \textbf{Final answer:} \) The resultant force is \(13\,\text{N}\).

Explanation: The centre of a regular hexagon is equidistant from all six vertices. For every vertex charge, there is an identical charge at the opposite vertex. The forces due to each opposite pair on the central \(+q\) are equal in magnitude and opposite in direction. These three pairs cancel separately. The net force is therefore zero because of symmetry. The important misconception warning is that many non-zero individual forces can still produce zero resultant when arranged symmetrically.

Explanation: Superposition is a vector method. Each source charge produces its own force on the selected charge according to Coulomb's law. The magnitude of each force depends on the charge product and separation, while its direction depends on charge signs and geometry. After calculating the individual force vectors, they must be added vectorially. Averaging charges or ignoring positions would lose the directional and distance information. The main mistake in superposition problems is treating a vector problem like a scalar arithmetic problem.

Explanation: \( \textbf{Given directions:} \) The \(x\)-component is along \(+x\), so it is \(+0.24\hat{i}\,\text{N}\). \( \textbf{Second component:} \) The \(y\)-component is along \(-y\), so it is \(-0.32\hat{j}\,\text{N}\). \( \textbf{Vector addition:} \) \[ \vec{F}_{\text{net}}=0.24\hat{i}\,\text{N}-0.32\hat{j}\,\text{N} \] \( \textbf{Compact form:} \) \[ \vec{F}_{\text{net}}=(0.24\hat{i}-0.32\hat{j})\,\text{N} \] \( \textbf{Sign meaning:} \) The negative sign before \(\hat{j}\) indicates direction along \(-y\), not a negative force magnitude. \( \textbf{Magnitude check if needed:} \) Its magnitude would be \(\sqrt{(0.24)^2+(0.32)^2}=0.40\,\text{N}\). \( \textbf{Final answer:} \) The correct vector form is \((0.24\hat{i}-0.32\hat{j})\,\text{N}\).

Explanation: A continuous charge distribution is a useful macroscopic model in which charge is treated as spread along a line, over a surface, or throughout a volume. This does not mean that microscopic charge quantisation becomes false. It only means that the number of elementary charges is so large that the distribution can be approximated as smooth. A continuous distribution can have positive, negative, or zero net charge. Coulomb interaction still exists, but the calculation often uses charge density and integration or symmetry. The key approximation is smoothness at ordinary scales, not denial of elementary charge.

Explanation: Linear charge density is charge per unit length, so its symbol is \(\lambda\) and its unit is \(\text{C m}^{-1}\). Surface charge density is charge per unit area, so its symbol is \(\sigma\) and its unit is \(\text{C m}^{-2}\). Volume charge density is charge per unit volume, so its symbol is \(\rho\) and its unit is \(\text{C m}^{-3}\). The negative powers of \(\text{m}\) show whether charge is divided by length, area, or volume. The most common symbol mistake is mixing up \(\lambda\), \(\sigma\), and \(\rho\). A quick unit check usually identifies the correct density.

Explanation: \( \textbf{Given:} \) Total charge \(Q=+12\,\mu\text{C}=+12\times10^{-6}\,\text{C}\), and rod length \(L=0.60\,\text{m}\). \( \textbf{Required:} \) Linear charge density \(\lambda\). \( \textbf{Formula:} \) \[ \lambda=\frac{Q}{L} \] \( \textbf{Why it applies:} \) The charge is uniformly spread along a rod, so charge per unit length is used. \( \textbf{Substitution:} \) \[ \lambda=\frac{12\times10^{-6}\,\text{C}}{0.60\,\text{m}} \] \( \textbf{Numerical simplification:} \) \[ \frac{12}{0.60}=20 \] \( \textbf{Scientific notation:} \) \[ \lambda=20\times10^{-6}\,\text{C m}^{-1}=2.0\times10^{-5}\,\text{C m}^{-1} \] \( \textbf{Sign check:} \) The charge is positive, so \(\lambda\) is positive. \( \textbf{Final answer:} \) The linear charge density is \(+2.0\times10^{-5}\,\text{C m}^{-1}\).