Electric Charges and Fields MCQs with Answers – Part 3 (Class 12 Physics)

Explanation: \( \textbf{Chosen sign convention:} \) Rightward force is positive. \( \textbf{First force:} \) \(F_1=+6\,\text{N}\). \( \textbf{Second force:} \) \(F_2=-10\,\text{N}\). \( \textbf{Superposition along a line:} \) \[ F_{\text{net}}=F_1+F_2 \] \( \textbf{Substitution:} \) \[ F_{\text{net}}=+6\,\text{N}-10\,\text{N} \] \( \textbf{Calculation:} \) \[ F_{\text{net}}=-4\,\text{N} \] \( \textbf{Direction meaning:} \) The negative sign means the resultant is toward the left. \( \textbf{Magnitude:} \) The magnitude is \(4\,\text{N}\), but the signed force is \(-4\,\text{N}\). \( \textbf{Final answer:} \) The net force is \(-4\,\text{N}\).

Explanation: The middle charge \(+q\) is repelled by the left \(+Q\) toward the right. It is also repelled by the right \(+Q\) toward the left. Since the two outer charges are equal and equally distant from the middle charge, the two force magnitudes are equal. Their directions are opposite along the same line. The resultant force is therefore zero. This is a collinear symmetry case where cancellation follows from equal magnitude and opposite direction.

Explanation: Both force contributions are stated to act toward \(+\hat{i}\). When vectors act along the same direction, their magnitudes add. The first contribution is \(0.90\,\text{N}\), and the second contribution is \(0.40\,\text{N}\). Therefore, the net magnitude is \(0.90\,\text{N}+0.40\,\text{N}=1.30\,\text{N}\). The direction remains \(+\hat{i}\) because neither contribution points left. Subtraction would be needed only if the two forces acted in opposite directions. In a collinear system, the sign of each force contribution must be assigned before adding.

Explanation: \( \textbf{Sign convention:} \) Right is \(+\hat{i}\), so left is \(-\hat{i}\). \( \textbf{Leftward force:} \) \[ \vec{F}_1=-0.72\hat{i}\,\text{N} \] \( \textbf{Rightward force:} \) \[ \vec{F}_2=+0.32\hat{i}\,\text{N} \] \( \textbf{Net-force relation:} \) \[ \vec{F}_{\text{net}}=\vec{F}_1+\vec{F}_2 \] \( \textbf{Substitution:} \) \[ \vec{F}_{\text{net}}=(-0.72+0.32)\hat{i}\,\text{N} \] \( \textbf{Calculation:} \) \[ \vec{F}_{\text{net}}=-0.40\hat{i}\,\text{N} \] \( \textbf{Direction meaning:} \) The negative sign means the resultant is toward left. \( \textbf{Final answer:} \) The vector net force is \(-0.40\hat{i}\,\text{N}\).

Explanation: The negative charge on the left attracts the middle positive charge, so that force is toward the left. The positive charge on the right repels the middle positive charge away from the right charge, which is also toward the left. Thus both contributions have the same direction. In this case, the net force is the sum of the two magnitudes toward the left. The result is not zero because the forces reinforce rather than cancel. A sign-position check is needed before deciding whether forces add or oppose.

Explanation: A zero net force requires vector cancellation. In a straight-line arrangement, this means force contributions must act in opposite directions and have equal magnitudes. If all contributions point in the same direction, they reinforce and cannot give zero resultant. The nearest charge still contributes force unless the charge value is zero or the setup provides another special condition. Coulomb’s law continues to depend on distance through \(\frac{1}{r^2}\). Locating a zero-force point is therefore a balance of directions and magnitudes, not just a search for a midpoint.

Explanation: Between two like positive charges, the small positive charge is repelled by both fixed charges in opposite directions. For cancellation, the stronger charge \(+4Q\) must be farther away than the weaker charge \(+Q\). Since Coulomb force varies as \(\frac{Q}{r^2}\) for the same test charge, distance can balance the larger charge. The zero-force point therefore lies closer to the smaller charge \(+Q\). The midpoint would not work because \(+4Q\) would exert the larger force there. Outside the two charges, the forces due to like charges on a positive test charge point in the same direction and do not cancel.

Explanation: The positive test charge is repelled by \(+Q\). Between the charges, repulsion from \(+Q\) pushes it away from \(+Q\), toward the negative charge. The negative charge \(-Q\) attracts the positive test charge, also toward \(-Q\). Thus the two electrostatic force contributions point in the same direction between unlike charges. They reinforce rather than cancel. Equal magnitudes would not produce zero there because their directions are not opposite.

Explanation: Row S is incorrect because a positive test charge midway between two equal positive charges is repelled by both. The left \(+Q\) pushes it toward the right, while the right \(+Q\) pushes it toward the left. These directions are opposite, not the same toward right. Row P correctly states the opposite-direction pattern between two like charges. Row Q is valid because between \(+Q\) on the left and \(-Q\) on the right, both forces point toward the right. Row R is the mirror case, so both forces point toward the left.

Explanation: Both fixed charges are positive, and the test charge is also positive. Each fixed charge repels the test charge. Since both fixed charges lie to the left of the test charge, repulsion from each one pushes the test charge toward the right. The two force contributions therefore act in the same direction. They do not cancel in the outside region for equal like charges. This contrasts with the region between equal like charges, where the two repulsive forces can oppose each other.

Explanation: If identical positive charges were placed at all four corners, the net force on the central positive charge would be zero by symmetry. With one corner empty, the force contribution that would have come from that missing corner is absent. The three existing corner charges repel the central charge. Their resultant is equivalent to the negative of the missing fourth contribution from the complete symmetric arrangement. Since a positive charge at the empty corner would have repelled the centre away from that corner, removing it leaves the resultant toward the empty corner. This is a symmetry-shortcut result, not a statement that the empty corner itself exerts force.

Explanation: \( \textbf{Forces involved:} \) The upper charge \(+q\) is repelled by both lower positive charges. \( \textbf{Magnitude of each contribution:} \) Each force has magnitude \(F\). \( \textbf{Angle between the two force vectors:} \) At the upper vertex of an equilateral triangle, the angle between the two sides is \(60^\circ\). \( \textbf{Resultant formula:} \) \[ R=\sqrt{F^2+F^2+2F^2\cos60^\circ} \] \( \textbf{Use \(\cos60^\circ=\frac{1}{2}\):} \) \[ R=\sqrt{F^2+F^2+F^2} \] \( \textbf{Simplification:} \) \[ R=\sqrt{3F^2} \] \( \textbf{Result:} \) \[ R=\sqrt{3}F \] \( \textbf{Direction note:} \) The horizontal components cancel and the vertical components add along the angle bisector. \( \textbf{Final answer:} \) The resultant force is \(\sqrt{3}F\).

Explanation: The two \(+Q\) charges are at opposite corners of one diagonal. They repel the central \(+q\) with equal magnitudes in opposite directions, so their forces cancel. The two \(-Q\) charges are also at opposite corners of the other diagonal. They attract the central \(+q\) with equal magnitudes in opposite directions, so those forces cancel as well. Since both diagonal pairs separately give zero resultant, the total force is zero. The signs of the charges matter for attraction or repulsion, but symmetry still controls cancellation at the centre.

Explanation: The charge at \((a,0)\) is positive, so it repels the positive charge at the origin toward the negative \(x\)-direction. That contribution is along \(-\hat{i}\). The charge at \((0,a)\) also repels the charge at the origin, but toward the negative \(y\)-direction. That contribution is along \(-\hat{j}\). Since the two fixed charges are identical and equally distant from the origin, the force magnitudes are equal. The resultant is therefore along the angle bisector of \(-\hat{i}\) and \(-\hat{j}\), namely along \(-\hat{i}-\hat{j}\).

Explanation: The two fixed charges are identical and symmetrically placed on the \(x\)-axis. The charge \(+q\) at \((0,a)\) is repelled by both fixed charges. The force due to the right charge has a leftward horizontal component, while the force due to the left charge has a rightward horizontal component. These horizontal components are equal in magnitude and opposite in direction. The vertical components both point upward and therefore add. Symmetry tells which components cancel and which components reinforce.

Explanation: \( \textbf{Force due to charge at \((a,0)\):} \) It is along \(-\hat{i}\) with magnitude \(F\). \( \textbf{Force due to charge at \((0,a)\):} \) It is along \(-\hat{j}\) with magnitude \(F\). \( \textbf{Angle between force contributions:} \) The two directions are perpendicular. \( \textbf{Vector addition for perpendicular forces:} \) \[ F_{\text{net}}=\sqrt{F^2+F^2} \] \( \textbf{Simplification:} \) \[ F_{\text{net}}=\sqrt{2F^2} \] \( \textbf{Result:} \) \[ F_{\text{net}}=\sqrt{2}F \] \( \textbf{Why \(2F\) is not used:} \) \(2F\) would apply only if both force vectors were in the same direction. \( \textbf{Final answer:} \) The net force magnitude is \(\sqrt{2}F\).

Explanation: The centre of a rectangle is the midpoint of each diagonal. The two fixed charges at opposite ends of one diagonal are therefore at equal distances from the central charge. Both are positive, so both repel the central \(+q\). The two repulsive forces lie along the same diagonal but in opposite directions. Equal magnitudes in opposite directions cancel. The cancellation depends on opposite-corner symmetry, not on the rectangle being a square.

Explanation: The two positive charges on the left repel the central positive charge away from the left side. Their vertical components cancel because the charges are symmetrically placed above and below the centre. Their horizontal components add toward the right. The two negative charges on the right attract the central positive charge toward the right side. Again, their vertical components cancel while their horizontal components add toward the right. Both the positive-charge pair and the negative-charge pair therefore reinforce a rightward resultant.

Explanation: The charges at top-left and bottom-right are both \(+Q\), placed at opposite corners. They repel the central \(+q\) with equal magnitudes in opposite directions along the same diagonal, so their forces cancel. The charges at top-right and bottom-left are both \(-Q\), also placed at opposite corners. They attract the central \(+q\) with equal magnitudes in opposite directions along the other diagonal, so their forces also cancel. The total vector sum is zero. Equal magnitude and opposite placement can cancel even when one diagonal contains positive charges and the other contains negative charges.

Explanation: \( \textbf{Geometry:} \) The two charges are at opposite corners of a square. \( \textbf{Square side:} \) \(a=0.20\,\text{m}\). \( \textbf{Diagonal relation:} \) \[ d=a\sqrt{2} \] \( \textbf{Substitution:} \) \[ d=0.20\sqrt{2}\,\text{m} \] \( \textbf{Meaning:} \) The separation in Coulomb’s law is the straight-line distance between the charges. \( \textbf{Why \(0.20\,\text{m}\) fails:} \) It is the side length, not the diagonal separation. \( \textbf{Why \(0.40\,\text{m}\) fails:} \) It adds two sides along a path instead of using direct separation. \( \textbf{Final answer:} \) The separation is \(0.20\sqrt{2}\,\text{m}\).