Q.Four point charges $q_A = 2\,\mu\text{C}$, $q_B = -5\,\mu\text{C}$, $q_C = 2\,\mu\text{C}$, and $q_D = -5\,\mu\text{C}$ are located at the corners of a square ABCD of side $10\,\text{cm}$. What is the force on a charge of $1\,\mu\text{C}$ placed at the centre of the square?

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Two identical balls each of mass $\dfrac{40}{3}$ g carry equal and opposite charge. They are suspended from a horizontal plate by silk threads each of length 1 m with a separation of 1.5 m between points of suspension. At equilibrium, if the distance between the balls is 30 cm then, magnitude of charge on each ball is (Acceleration due to gravity $= 10\ ms^{-2}$) (A) $10^{-6}\ C$ (B) $10^{-5}\ C$ (C) $10^{-4}\ C$ (D) $10^{-3}\ C$

   ›Reveal solutionSolution

   Two oppositely charged balls hung from points 1.5 m apart pull together to a 30 cm separation; equating the horizontal Coulomb attraction to $mg\tan\theta$ gives $q\approx 10^{-6}$ C.

   Concept and Intuition

   Each ball hangs on a 1 m thread. If uncharged, gravity alone would make each thread hang straight down, so the two balls would sit directly below their suspension points — meaning the natural, force-free separation between the balls equals the 1.5 m separation between the suspension points. Because the two charges are opposite in sign, the Coulomb force between them is attractive, so the balls are pulled towards each other until they settle at the smaller observed separation of 30 cm. At equilibrium, each thread makes an angle $\theta$ with the vertical, and the horizontal component of tension balances the attractive electric force while the vertical component balances gravity — the classic "tan θ" trick used for suspended-charge problems.

   Step-by-Step Solution

   1. Mass of each ball: $m = \dfrac{40}{3}\times 10^{-3}\ \text{kg}$, so $mg = \dfrac{40}{3}\times10^{-3}\times10 = 0.1\overline{3}\ \text{N}$.
   2. Natural separation (uncharged) = suspension-point separation = 1.5 m. Actual separation = 0.30 m, so total inward shift = $1.5-0.3 = 1.2$ m, i.e. each ball/thread moves in by $0.6$ m.
   3. Each thread has length 1 m and horizontal displacement 0.6 m: $\sin\theta = 0.6/1 = 0.6$, so $\cos\theta = 0.8$ and $\tan\theta = 0.6/0.8 = 0.75$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If A, B and C represent the work done, distance and electric charge respectively, Then the physical quantity having the dimensions of $\frac{C^2}{AB}$ is (A) Permittivity (B) Permeability (C) Electric potential (D) Electric energy

   ›Reveal solutionSolution

   Working out the dimensional formula of $C^2/(AB)$ (charge² over work·distance) shows it matches the dimensional formula of the permittivity of free space, $\varepsilon_0$.

   Concept and Intuition

   Coulomb's law $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ directly ties $\varepsilon_0$ to charge, force, and distance: $\varepsilon_0=\dfrac{q^2}{4\pi F r^2}$. Since force $=$ work/distance, this becomes $\varepsilon_0\propto \dfrac{\text{charge}^2}{(\text{work})(\text{distance})}$ — exactly the combination $C^2/(AB)$ given in the problem (with $A=$ work, $B=$ distance, $C=$ charge).

   Step-by-Step Solution

   1. Write the dimensional formulas: $[A]=\text{work}=ML^2T^{-2}$, $[B]=\text{distance}=L$, $[C]=\text{charge}=IT$.
   2. Compute $\dfrac{C^2}{AB}=\dfrac{(IT)^2}{(ML^2T^{-2})(L)}=\dfrac{I^2T^2}{ML^3T^{-2}}=I^2T^4M^{-1}L^{-3}$.
   3. Recall $\varepsilon_0$ from Coulomb's law: $\varepsilon_0=\dfrac{q_1q_2}{4\pi F r^2}$, so $[\varepsilon_0]=\dfrac{(IT)^2}{(MLT^{-2})(L^2)}=\dfrac{I^2T^2}{ML^3T^{-2}}=I^2T^4M^{-1}L^{-3}$.
   4. This exactly matches the dimensional formula computed for $C^2/(AB)$ in step 2. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Charges q, 2q, 3q and 4q are placed at the corners A, B, C and D of a square as shown in the figure. The direction of electric field at the centre 'O' of the square is along [FIGURE: a square with A at the bottom-left carrying charge $q$, B at the bottom-right carrying charge $2q$, C at the top-right carrying charge $3q$, D at the top-left carrying charge $4q$; O is the centre of the square; the diagonals AC and BD are drawn dashed, intersecting at O] (A) AB (B) CB (C) BD (D) AC

   ›Reveal solutionSolution

   Since all four corners are equidistant from the centre, each charge contributes a field of magnitude $\propto$ its charge, directed away from itself through $O$. Vector-adding all four (or pairing diagonals first) gives a resultant pointing along CB.

   Concept and Intuition

   At the centre of a square, every corner is the same distance from $O$ (half the diagonal), so the relative field contributions depend only on the charge magnitudes, each directed radially outward from its own corner (for positive charges) through $O$. The cleanest way to combine four such vectors is to pair the diagonally-opposite ones first, since their directions are exactly opposite (anti-parallel) along each diagonal — the pair simply nets to (larger charge $-$ smaller charge) pointing away from the larger one, i.e. toward the smaller-charge corner.

   Step-by-Step Solution

   1. Assign coordinates matching the figure: $A(0,0)=q$, $B(1,0)=2q$, $C(1,1)=3q$, $D(0,1)=4q$, centre $O(0.5,0.5)$.
   2. Diagonal AC ($A=q$, $C=3q$): the two fields point in opposite directions along AC. Net magnitude $\propto |q-3q|=2q$, directed from $C$ toward $A$ (pointing away from the larger charge, $C$).
   3. Diagonal BD ($B=2q$, $D=4q$): similarly, net magnitude $\propto |2q-4q|=2q$, directed from $D$ toward $B$ (away from the larger charge, $D$).
   4. Both diagonals are perpendicular to each other, so resolve into $x$–$y$ components and add:

   • AC-pair vector: magnitude $2q$ along direction $(-1,-1)/\sqrt2$ (toward $A$): components $(-\sqrt2 q,-\sqrt2 q)$.
   • BD-pair vector: magnitude $2q$ along direction $(1,-1)/\sqrt2$ (toward $B$): components $(\sqrt2 q,-\sqrt2 q)$. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Two point charges Q and $-4Q$ are separated by a distance r. If electric field at the location of Q is $\vec{E}$, then the field at the location of $-4Q$ is (A) $\vec{E}$ (B) $-\vec{E}$ (C) $-\dfrac{\vec{E}}{4}$ (D) $\dfrac{\vec{E}}{4}$

   ›Reveal solutionSolution

   This tests careful use of Coulomb's law with vector direction: the field due to one point charge at the other charge's location is found by considering only the source charge's magnitude and sign, and both fields here point the same way along the line joining the charges.

   Concept and Intuition

   The electric field "at the location of $Q$" must be produced by the other charge, $-4Q$ (a charge doesn't act on itself). Likewise, the field "at the location of $-4Q$" is produced only by $Q$. Both fields lie along the line joining the two charges, so we just need their magnitudes (Coulomb's law, $E = kq/r^2$) and their directions (field points away from a positive source, toward a negative source).

   Step-by-Step Solution

   1. Set up coordinates: put $Q$ at the origin and $-4Q$ at position $x=r$ (so the vector from $Q$ to $-4Q$ points in $+\hat{x}$).
   2. Field at $Q$'s location due to $-4Q$: since $-4Q$ is negative, the field it creates points toward itself — i.e. from $Q$'s position toward $-4Q$'s position, which is the $+\hat x$ direction. Magnitude $= \dfrac{k(4Q)}{r^2}$. So $\vec{E} = \frac{4kQ}{r^2}\,\hat{x}.$
   3. Field at $-4Q$'s location due to $Q$: since $Q$ is positive, the field it creates points away from itself — i.e. also in the $+\hat x$ direction (away from $Q$, through $-4Q$'s position and beyond). Magnitude $=\dfrac{kQ}{r^2}$. So this field $= \dfrac{kQ}{r^2}\hat x$. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If two positive charges each of 20 $\mu$C are placed at the two vertices of an equilateral triangle of side 50 cm and a third positive charge of $10\sqrt{3}$ $\mu$C is placed at the centroid of the triangle, then the electrostatic potential energy of the system of three charges is (A) 14.4 J (B) 57.6 J (C) 28.8 J (D) 21.6 J

   ›Reveal solutionSolution

   This tests computing total electrostatic potential energy of a 3-charge system by summing all three pairwise interaction energies, using the geometric fact that an equilateral triangle's centroid is at distance $a/\sqrt3$ from each vertex.

   Concept and Intuition

   The total electrostatic potential energy of a system of point charges is the sum of the potential energies of every distinct pair:

   $U = \sum_{i<j}\frac{kq_iq_j}{r_{ij}}.$

   Here two charges sit at vertices of an equilateral triangle (separated by the full side length $a$), and the third sits at the centroid. For an equilateral triangle of side $a$, the distance from the centroid to each vertex is the circumradius, $R = \dfrac{a}{\sqrt3}$ (derivable from height $h=\frac{\sqrt3}{2}a$ and centroid dividing the median in ratio $2:1$ from the vertex, giving $R=\frac23 h = \frac{a}{\sqrt3}$).

   Step-by-Step Solution

   1. Side of triangle: $a = 50\ \text{cm} = 0.5\ \text{m}$.
   2. Charges at the two vertices: $q_1=q_2=20\ \mu\text{C} = 20\times10^{-6}$ C. Charge at centroid: $q_3 = 10\sqrt3\ \mu\text{C} = 10\sqrt3\times10^{-6}$ C.
   3. Distance between the two vertex charges: $r_{12} = a = 0.5$ m.
   4. Distance from centroid to each vertex: $r_{13}=r_{23} = \dfrac{a}{\sqrt3} = \dfrac{0.5}{\sqrt3}$ m.
   5. Pairwise energy between the two vertex charges:

   $U_{12} = \frac{kq_1q_2}{r_{12}} = \frac{9\times10^9\times(20\times10^{-6})^2}{0.5} = \frac{9\times10^9\times4\times10^{-10}}{0.5} = \frac{3.6}{0.5} = 7.2\ \text{J}.$

   6. Pairwise energy between a vertex charge and the centroid charge: $U_{13} = \frac{kq_1q_3}{a/\sqrt3} = \frac{9\times10^9\times20\times10^{-6}\times10\sqrt3\times10^{-6}}{0.5/\sqrt3}.$ …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Two similar rods of length $l=1m$ carrying equal charges $(q)=10^{-8}C$ are placed as shown in figure. The electric field at point 'O' approximately is, if $d=0.25m$ [FIGURE: one rod is vertical carrying positive charge, standing at height $l$ above point O with a gap d between the rod's lower end and O; a second rod lies horizontal carrying positive charge, starting at horizontal distance d from O and extending length $l$ to the right] (A) $450\ Vm^{-1}$ (B) $568\ Vm^{-1}$ (C) $406\ Vm^{-1}$ (D) $203\ Vm^{-1}$

   ›Reveal solutionSolution

   Each rod produces an axial field at O of magnitude $k\lambda l/[d(d+l)]$, directed away from it; the two contributions are mutually perpendicular (one vertical, one horizontal) and equal in magnitude, so the resultant is $E_0\sqrt2\approx406\ Vm^{-1}$.

   Concept and Intuition

   For a point lying on the extension of a uniformly charged rod's own axis (i.e. off one end, along the line containing the rod), the field is found by integrating Coulomb's law element-by-element along the rod: $E=\displaystyle\int_d^{d+l}\dfrac{k\lambda\,dy}{y^2}=k\lambda\left(\dfrac1d-\dfrac1{d+l}\right)=\dfrac{k\lambda l}{d(d+l)}$. Here, O sits exactly on the axis of each rod (the vertical rod's axis is the vertical line through O; the horizontal rod's axis is the horizontal line through O), so this formula applies twice — once for each rod — and the two resulting field vectors are perpendicular because the rods themselves are perpendicular. Equal charge, length, and gap $d$ make the two magnitudes identical, so they combine via Pythagoras.

   Step-by-Step Solution

   1. Linear charge density: $\lambda=\dfrac{q}{l}=\dfrac{10^{-8}}{1}=10^{-8}$ C/m.
   2. Axial field magnitude from one rod: $E_0=\dfrac{k\lambda l}{d(d+l)}$, with $k=9\times10^9$, $d=0.25$ m, $l=1$ m so $d+l=1.25$ m.
   3. $E_0=\dfrac{9\times10^9\times10^{-8}\times1}{0.25\times1.25}=\dfrac{90}{0.3125}=288$ V/m. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The force between two point charges kept with a separation of 9 cm in air is 98 N. If a dielectric slab of constant 4, thickness 6 cm and another dielectric slab of constant 9, thickness 3 cm are introduced between the two charges, then the new force becomes (A) 18 N (B) 36 N (C) 49 N (D) 84 N

   ›Reveal solutionSolution

   This tests the effect of dielectric slabs placed between two point charges on the Coulomb force between them, using the effective-separation method. Answer: 18 N.

   Concept and Intuition

   When the entire gap between two point charges is filled with a dielectric of constant $K$, the force reduces by a factor of $K$ (Coulomb's law inside a dielectric medium). When only part of the path is filled by slabs of finite thickness, each slab behaves as though it stretches that portion of the path by a factor $\sqrt{K}$ — exactly as an optical medium of refractive index $n=\sqrt{K}$ increases the effective (optical) path length of light travelling through it. So the total effective separation becomes the sum of the untouched air gaps plus $t\sqrt{K}$ for each slab, and the force follows the inverse-square law using this effective separation.

   Step-by-Step Solution

   1. Original separation $d=9$ cm, giving $F=98$ N in air.
   2. Slab 1: thickness $t_1=6$ cm, constant $K_1=4$, so $\sqrt{K_1}=2$; its effective length becomes $t_1\sqrt{K_1}=12$ cm.
   3. Slab 2: thickness $t_2=3$ cm, constant $K_2=9$, so $\sqrt{K_2}=3$; its effective length becomes $t_2\sqrt{K_2}=9$ cm.
   4. Since $t_1+t_2=6+3=9$ cm $=d$, the slabs fill the whole gap (no leftover air gap). …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Three particles of each charge q are placed at the vertices of an equilateral triangle of side L. The work to be done to decrease the side of the triangle to $\dfrac{L}{2}$ is (A) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{L}$ (B) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q^2}{L}$ (C) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{3q^2}{L}$ (D) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{3q^2}{2L}$

   ›Reveal solutionSolution

   The work needed to squeeze the triangle equals the increase in total electrostatic potential energy of the three like charges, which comes out to $\dfrac{1}{4\pi\varepsilon_0}\dfrac{3q^2}{L}$.

   Concept and Intuition

   By the work-energy theorem for conservative electrostatic forces, the external work done to slowly rearrange a system of charges (with no change in kinetic energy) equals the change in the system's total electrostatic potential energy. For like charges (here all charge $q$, presumably same sign since they repel and require positive work to bring closer), moving them closer together requires positive work because we're pushing against their mutual repulsion.

   Step-by-Step Solution

   1. For a system of point charges, total PE is the sum over all distinct pairs: $U = \sum \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_iq_j}{r_{ij}}$.
   2. In an equilateral triangle of side $L$ with three equal charges $q$, there are 3 pairs, each at separation $L$: $U_i = 3\times\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{L}$.
   3. When the side shrinks to $L/2$ (still equilateral, so all three pairwise distances are equal to $L/2$): $U_f = 3\times\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{L/2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{6q^2}{L}$. …

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9. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.Three point charges $+10\ \mu$C, $+20\ \mu$C and $+40\ \mu$C are placed at the vertices of an equilateral triangle of side 1m. The electrostatic potential energy of the system of the charges is (A) 6.3 J (B) 12.6 J (C) 25.2 J (D) 21.6 J

   ›Reveal solutionSolution

   Total electrostatic potential energy of a system of point charges is the sum over every pair; for this equilateral triangle (all sides 1 m) it works out to 12.6 J.

   Concept and Intuition

   Potential energy of a system of charges is the work needed to assemble them, which is the sum of the pairwise interaction energies $\dfrac{kq_iq_j}{r_{ij}}$ over all distinct pairs. Since the triangle is equilateral, every pairwise distance is the same (1 m), which simplifies the calculation to a single sum of products.

   Step-by-Step Solution

   1. Pairs and their charge products: $q_1q_2=10\times20=200$, $q_1q_3=10\times40=400$, $q_2q_3=20\times40=800$ (all in $\mu C^2$).
   2. Sum: $200+400+800=1400\ \mu C^2 = 1400\times10^{-12}\ C^2$. …

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10. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.If three particles of each charge $+q$ are placed at the three vertices of an equilateral triangle of side $\sqrt{3}\,r$, then the net electric field at the centroid of the triangle is (A) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ (B) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ (C) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{3q}{r^2}$ (D) zero

    ›Reveal solutionSolution

    Three identical charges at the vertices of an equilateral triangle create fields at the centroid that are equal in magnitude and symmetrically arranged at $120^\circ$ to each other — they cancel exactly, giving zero net field.

    Concept and Intuition

    The centroid of an equilateral triangle is equidistant from all three vertices, so each charge contributes an electric field of the same magnitude at that point. Because the triangle has three-fold rotational symmetry, the three field vectors (each pointing away from its source charge, since the charges are positive) are oriented $120^\circ$ apart from one another. Three equal-magnitude vectors symmetrically spaced by $120^\circ$ always add to zero — this is a standard vector identity (same reasoning as why three equal forces at $120^\circ$ balance).

    Step-by-Step Solution

    1. Side of triangle $= \sqrt{3}\,r$; centroid-to-vertex distance for an equilateral triangle of side $a$ is $\dfrac{a}{\sqrt3}$.
    2. So centroid-to-vertex distance $= \dfrac{\sqrt3\,r}{\sqrt3} = r$.
    3. Each charge contributes field magnitude $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ at the centroid, directed away from that vertex. …

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11. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.Three point charges +Q, q and +Q are placed on x-axis at distances $0, \dfrac{d}{2}$ and d respectively from the origin. If the resultant electrostatic force on the point charge +Q placed at $x = 0$ is zero, then the value of q is (A) $+\dfrac{Q}{2}$ (B) $-\dfrac{Q}{2}$ (C) $-\dfrac{Q}{4}$ (D) $+\dfrac{Q}{4}$

    ›Reveal solutionSolution

    This tests force balance for three collinear charges; the middle charge must be negative and of magnitude Q/4 to cancel the repulsion from the far +Q.

    Concept and Intuition

    The charge +Q at the origin feels two forces: a repulsion from the +Q at $x=d$ (pushing it toward $-x$), and a force from $q$ at $x=d/2$. For the net force to be zero, the force from $q$ must point in the $+x$ direction — i.e., $q$ must attract the origin charge, meaning $q$ is negative.

    Step-by-Step Solution

    1. Force from +Q at x=d on the origin charge: $F_2 = \dfrac{kQ^2}{d^2}$, directed along $-x$ (repulsive).
    2. Force from q at $x=d/2$ on origin charge must be equal in magnitude and directed along $+x$ (attractive), so q < 0.
    3. Magnitude: $F_1 = \dfrac{kQ|q|}{(d/2)^2} = \dfrac{4kQ|q|}{d^2}$. …

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12. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The sum of two point positive charges separated by a distance of 1.5 m in air is $25 \mu\mathrm{C}$. If the electrostatic force between the two charges is 0.6 N, then the difference between the two charges is (A) $5 \mu\mathrm{C}$ (B) $8 \mu\mathrm{C}$ (C) $3 \mu\mathrm{C}$ (D) $6 \mu\mathrm{C}$

    ›Reveal solutionSolution

    This tests using Coulomb's law together with a sum constraint to find two unknown charges. The difference between the charges comes out to $5\,\mu\mathrm{C}$.

    Concept and Intuition

    Coulomb's law gives the product of two charges from a measured force, while the problem separately gives their sum. Two numbers with a known sum and a known product are exactly the roots of a quadratic equation — this is the standard algebraic trick used whenever 'sum and product' of two charges (or masses, etc.) are both known.

    Step-by-Step Solution

    1. Coulomb's law: $F = \dfrac{k\,q_1q_2}{r^2}$, so $q_1q_2 = \dfrac{F r^2}{k} = \dfrac{0.6 \times (1.5)^2}{9\times10^9} = \dfrac{0.6\times 2.25}{9\times10^9} = 1.5\times10^{-10}\ \mathrm{C^2}$.
    2. Convert to $\mu$C$^2$: $1.5\times10^{-10}\ \mathrm{C^2} = 150\ (\mu\mathrm{C})^2$ (since $1\,(\mu\mathrm C)^2 = 10^{-12}\,\mathrm C^2$).
    3. Given $q_1+q_2 = 25\,\mu$C and $q_1 q_2 = 150\,(\mu\mathrm{C})^2$, form the quadratic $x^2 - 25x + 150 = 0$. …

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