Q.Two point charges, each equal to $-q$, are held fixed on a straight line, separated by a distance $2d$ (so each is a distance $d$ from the mid-point of the line). A third charge $+q$ of mass $m$ is placed at the mid-point and is then displaced by a small distance $x$ (with $x \ll d$) in the direction perpendicular to the line joining the two fixed charges. Show that the charge $+q$ executes simple harmonic motion, and that its time period is $T = \left[\dfrac{8\pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1/2}$.

Showing the 12 most recent of 31 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The shift (in metre) in center of mass when the largest possible equilateral triangular plate is removed from a uniform square plate of side 2 m with one of their sides coinciding is (A) $\dfrac{\sqrt{3}-4}{\sqrt{3}-2}$ (B) $\dfrac{(\sqrt{3}-1)}{(4-\sqrt{3})}$ (C) $\dfrac{2-\sqrt{3}}{4-\sqrt{3}}$ (D) $\dfrac{\sqrt{3}-1}{\sqrt{3}-4}$

   ›Reveal solutionSolution

   Removing the equilateral triangle (area $\sqrt3$) from the square (area $4$) shifts the CM by $\dfrac{\sqrt3-1}{4-\sqrt3}$ m.

   The largest equilateral triangle whose base coincides with a side of the square has side $=2$ m (its height $\sqrt3\approx1.732<2$, so it fits). Take the square as $[0,2]\times[0,2]$ with the coinciding side along $y=0$; the triangle has apex at $(1,\sqrt3)$.

   Treat area as mass ($\sigma=1$):

   • Square: $M_1=4$, centre $y_1=1$.
   • Triangle: $M_2=\dfrac{\sqrt3}{4}(2)^2=\sqrt3$, centroid $y_2=\dfrac{\sqrt3}{3}$.

   By symmetry the CM stays at $x=1$; the $y$-coordinate of the remaining plate: …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.In an isosceles right angled triangle ABC, the length of the two equal sides AB and AC is 10 cm. If three charges +5 $\mu$C, +20 $\mu$C and +20 $\mu$C are placed at the three vertices A, B and C of the triangle respectively, then the net electrostatic force acting on a particle of charge +2 $\mu$C placed at the midpoint of the hypotenuse BC is (A) 24 N (B) 9 N (C) 36 N (D) 18 N

   ›Reveal solutionSolution

   The net force on the +2 μC charge at the midpoint of BC is the vector sum of the repulsive forces from A, B, and C. Due to symmetry, the forces from B and C cancel in the horizontal direction and add vertically, while the force from A points directly away from A. The result is 18 N, corresponding to option (D).

   Concept and Intuition

   The problem is a classic application of Coulomb’s law and vector addition. The key insight is that the midpoint of the hypotenuse in an isosceles right triangle is equidistant from B and C, and also has a special geometric relationship with vertex A. Because charges at B and C are equal (+20 μC each), the forces they exert on the test charge are symmetric. This symmetry simplifies the vector sum: the horizontal components cancel, and the vertical components add. The force from A is purely along the line from A to the midpoint. Adding these contributions gives the net force.

   Step-by-step solution

   1. Set up coordinates and distances Place the triangle with right angle at A. Let A = (0,0), B = (10,0) cm, C = (0,10) cm. The hypotenuse BC runs from (10,0) to (0,10). Its midpoint M is at

   $M = \left( \frac{10+0}{2}, \frac{0+10}{2} \right) = (5,5) \text{ cm}.$

   Distances:

   • From A to M: $AM = \sqrt{5^2 + 5^2} = 5\sqrt{2} \text{ cm} = 0.05\sqrt{2} \text{ m}.$
   • From B to M: $BM = \sqrt{(10-5)^2 + (0-5)^2} = \sqrt{25+25} = 5\sqrt{2} \text{ cm} = 0.05\sqrt{2} \text{ m}.$
   • Similarly, $CM = 5\sqrt{2} \text{ cm}$. So all three distances are equal: $r = 0.05\sqrt{2}$ m.

   2. Compute individual forces using Coulomb’s law Coulomb’s law: $F = k \frac{|q_1 q_2|}{r^2}$, with $k = 9 \times 10^9 \ \text{N m}^2/\text{C}^2$.
      • Force from A:

   $F_A = 9 \times 10^9 \times \frac{(5 \times 10^{-6})(2 \times 10^{-6})}{(0.05\sqrt{2})^2} = 9 \times 10^9 \times \frac{10 \times 10^{-12}}{0.005} = 9 \times 10^9 \times 2 \times 10^{-9} = 18 \ \text{N}.$

   • Force from B:

   $F_B = 9 \times 10^9 \times \frac{(20 \times 10^{-6})(2 \times 10^{-6})}{(0.05\sqrt{2})^2} = 9 \times 10^9 \times \frac{40 \times 10^{-12}}{0.005} = 9 \times 10^9 \times 8 \times 10^{-9} = 72 \ \text{N}.$

   • Force from C is identical: $F_C = 72 \ \text{N}$.

   3. Determine directions

      All charges are positive, so forces are repulsive.

      • $\vec{F}_A$ points from M away from A. Vector from A to M is (5,5), so direction is along the line $y = x$ away from origin. Unit vector: $\hat{u}_A = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$.
      • $\vec{F}_B$ points from M away from B. B is at (10,0), M at (5,5), so vector from B to M is (-5,5). Direction from M away from B is opposite: (5,-5). Unit vector: $\hat{u}_B = \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$.
      • $\vec{F}_C$ points from M away from C. C is at (0,10), M at (5,5), so vector from C to M is (5,-5). Direction away from C is (-5,5). Unit vector: $\hat{u}_C = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$.

   4. Vector addition

      Write components:

   $\vec{F}_A = 18 \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{18}{\sqrt{2}}, \frac{18}{\sqrt{2}} \right)$

   $\vec{F}_B = 72 \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) = \left( \frac{72}{\sqrt{2}}, -\frac{72}{\sqrt{2}} \right)$

   $$$$

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Four particles P, Q, R and S of masses m, m, m and 2m respectively are kept at the four corners of a square of side $\sqrt{2}$ m. The distance of centre of mass of the system of particles from the particle S is (A) 1.2 m (B) 0.8 m (C) 0.6 m (D) 0.4 m

   ›Reveal solutionSolution

   Place the square in a coordinate system, compute the centre of mass using the weighted average of positions, then find its distance from the corner where S (mass $2m$) sits. The answer is 0.4 m.

   The centre of mass of a system of particles is the point where the entire mass can be thought to be concentrated for translational motion. It is found by taking a weighted average of the positions, with each mass as the weight. For a discrete set, the formula is:

   $\vec{R}_{\text{CM}} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$

   Here we have four particles at the corners of a square. The trick is to choose a convenient coordinate system so that the arithmetic is clean. Since the side length is $\sqrt{2}$ m, placing the square with its sides parallel to the axes makes the coordinates simple.

   1. Set up coordinates. Let the square have side $\sqrt{2}$ m. Place corner S at the origin $(0,0)$. Then the other corners can be placed as:

      • P at $(\sqrt{2}, 0)$
      • Q at $(\sqrt{2}, \sqrt{2})$
      • R at $(0, \sqrt{2})$
      • S at $(0, 0)$

      This is a square of side $\sqrt{2}$ m, as required.

   2. List masses and positions.

      • P: mass $m$, position $(\sqrt{2}, 0)$
      • Q: mass $m$, position $(\sqrt{2}, \sqrt{2})$
      • R: mass $m$, position $(0, \sqrt{2})$
      • S: mass $2m$, position $(0, 0)$

   3. Compute total mass.

   $M = m + m + m + 2m = 5m$

   4. Find the $x$-coordinate of the centre of mass.

   $x_{\text{CM}} = \frac{m(\sqrt{2}) + m(\sqrt{2}) + m(0) + 2m(0)}{5m} = \frac{2m\sqrt{2}}{5m} = \frac{2\sqrt{2}}{5}$

   5. Find the $y$-coordinate of the centre of mass. $y_{\text{CM}} = \frac{m(0) + m(\sqrt{2}) + m(\sqrt{2}) + 2m(0)}{5m} = \frac{2m\sqrt{2}}{5m} = \frac{2\sqrt{2}}{5}$ …

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.If three electric charges each of magnitude $20~\mu \mathrm{C}$ are placed at any three corners of a square of side $\sqrt{2}~\mathrm{m}$, then the net electric field at the centre of the square (in $10^5~\mathrm{NC}^{-1}$) is (A) 1.2 (B) 5.4 (C) 3.6 (D) 1.8

   ›Reveal solutionSolution

   The centre lies 1 m from each corner; the two charges on a diagonal cancel, leaving the field of the single unpaired charge $= 1.8\times10^5~\text{NC}^{-1}$.

   Geometry. For a square of side $a=\sqrt{2}~\text{m}$, the diagonal is $a\sqrt{2}=2~\text{m}$, so the distance from the centre to each corner is

   $r=\frac{\text{diagonal}}{2}=1~\text{m}.$

   Field of one charge at the centre.

   $E=\frac{kq}{r^{2}}=\frac{(9\times10^{9})(20\times10^{-6})}{(1)^{2}}=1.8\times10^{5}~\text{NC}^{-1}.$ …

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5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Three charges q, q and Q (q = +20 $\mu$C and Q = +10 $\mu$C) are placed on the circumference of a circle of radius 10$\sqrt{3}$ cm. If the distance between any two charges is same, then the total electrostatic potential energy of the system of the three charges is (A) 48 J (B) 36 J (C) 24 J (D) 12 J

   ›Reveal solutionSolution

   The three charges form an equilateral triangle. The total electrostatic potential energy is the sum of the potential energies of all unique pairs of charges, which calculates to $\boxed{24 \text{ J}}$.

   The electrostatic potential energy of a system of charges represents the total work done by an external agent to assemble these charges from infinity to their current positions. This work is stored as potential energy in the system.

   For a system of multiple point charges, the total electrostatic potential energy is the algebraic sum of the potential energies of all unique pairs of charges. We consider each pair of charges independently and sum their individual potential energies.

   The electrostatic potential energy $U$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by:

   $U = \frac{k q_1 q_2}{r}$

   where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant, approximately $9 \times 10^9 \text{ N m}^2/\text{C}^2$.

   Here's how to calculate the total potential energy for the given system:

   1. Determine the geometry and distance between charges:

      The problem states that three charges are placed on the circumference of a circle of radius $R = 10\sqrt{3}$ cm, and the distance between any two charges is the same. This configuration implies that the charges form an equilateral triangle inscribed within the circle.

      For an equilateral triangle inscribed in a circle of radius $R$, the side length $a$ (which is the distance between any two charges) is related to the radius by the formula $a = R\sqrt{3}$.

      Let's convert the radius to meters:

      $R = 10\sqrt{3} \text{ cm} = 10\sqrt{3} \times 10^{-2} \text{ m} = \sqrt{3} \times 10^{-1} \text{ m}$.

      Now, calculate the side length $a$:

      $a = R\sqrt{3} = (\sqrt{3} \times 10^{-1} \text{ m}) \times \sqrt{3} = 3 \times 10^{-1} \text{ m} = 0.3 \text{ m}$.

      So, the distance between any two charges is $0.3 \text{ m}$.

   2. Identify all unique pairs of charges:

      We have three charges: $q_1 = q = +20 \mu\text{C}$, $q_2 = q = +20 \mu\text{C}$, and $q_3 = Q = +10 \mu\text{C}$.

      There are three unique pairs in a system of three charges:

      • Pair 1: $(q_1, q_2)$ which are $(q, q)$
      • Pair 2: $(q_1, q_3)$ which are $(q, Q)$
      • Pair 3: $(q_2, q_3)$ which are $(q, Q)$

   3. Apply the formula for total potential energy:

      The total electrostatic potential energy $U_{total}$ is the sum of the potential energies of these three pairs:

      $U_{total} = U_{12} + U_{13} + U_{23}$

      Since the distance between any two charges is the same ($a = 0.3 \text{ m}$), we can write: …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A charge q is placed at the centre ‘O’ of a circle of radius R and two other charges q and q are placed at the ends of the diameter AB of the circle. The work done to move the charge at point B along the circumference of the circle to a point C as shown in the figure is (A) $\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{R} (\sqrt{2})$ (B) Zero (C) $\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{R} \left(\dfrac{\sqrt{2}-1}{2}\right)$ (D) $\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{R} \left(\sqrt{\dfrac{1}{2}}\right)$

   ›Reveal solutionSolution

   The work done equals the change in electrostatic potential energy of the system. Since the charge at B moves along an equipotential of the central charge, only the interaction with the other fixed charge at A changes. The result is $\frac{1}{4\pi\epsilon_0} \frac{q^2}{R} \left(\frac{\sqrt{2}-1}{2}\right)$, which is option (C).

   Concept and intuition:

   Work done by an external agent to move a charge slowly (without gaining kinetic energy) equals the change in the system’s electrostatic potential energy. Here, we have three charges: one fixed at the centre O, one fixed at A, and one moved from B to C along the circle. The central charge creates a potential that is constant on the circle (all points are at distance $R$ from O), so moving along the circle does no work against that central charge. The only change comes from the interaction between the moving charge and the fixed charge at A, because the distance between them changes from $2R$ (diameter) to $\sqrt{2}R$ (chord of a right angle).

   ---

   Step-by-step solution:

   1. Identify the initial and final configurations.

      Initially, the three charges are:

      • $+q$ at centre O
      • $+q$ at A (one end of diameter)
      • $+q$ at B (other end of diameter) Finally, the charge at B moves to point C, which is on the circumference such that $\angle AOC = 90^\circ$ (since C is at the end of a perpendicular radius). So triangle AOC is right-angled at O, with OA = OC = $R$, hence AC = $\sqrt{R^2 + R^2} = \sqrt{2}R$.

   2. Work done = change in potential energy.

      The work done by an external agent is $W = U_{\text{final}} - U_{\text{initial}}$, where $U$ is the total electrostatic potential energy of the system of three point charges.

   3. Write the general formula for potential energy.

      For three charges $q_1, q_2, q_3$ at pairwise distances $r_{12}, r_{13}, r_{23}$:

   $U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right).$

   4. Compute initial potential energy $U_i$.
      • Distance O–A = $R$, O–B = $R$, A–B = $2R$.
      • All charges are $+q$.

   $U_i = \frac{1}{4\pi\epsilon_0} \left( \frac{q^2}{R} + \frac{q^2}{R} + \frac{q^2}{2R} \right) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{R} \left( 2 + \frac{1}{2} \right) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{R} \cdot \frac{5}{2}.$

   5. Compute final potential energy $U_f$. After moving the charge from B to C:
      • O–A = $R$ (unchanged)
      • O–C = $R$ (still on the circle)
      • A–C = $\sqrt{2}R$

   $U_f = \frac{1}{4\pi\epsilon_0} \left( \frac{q^2}{R} + \frac{q^2}{R} + \frac{q^2}{\sqrt{2}R} \right) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{R} \left( 2 + \frac{1}{\sqrt{2}} \right).$

   6. Find the work done. …

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7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.When three particles each having a positive charge ‘q’ are placed at the three vertices of an equilateral triangle, then the electrostatic force between any two particles is F. If a fourth particle of charge ‘3q’ is placed at the midpoint of one of the sides of the triangle, then the net electrostatic force on the fourth particle due to the remaining three particles is (A) $3F$ (B) $\sqrt{3}F$ (C) $4F$ (D) $9F$

   ›Reveal solutionSolution

   The two charges on the same side lie on a straight line through the fourth charge, so their forces are equal and opposite and cancel. Only the vertex charge opposite that side contributes, giving a net force of $4F$ (option C).

   Set-up. Three charges $q$ sit at the vertices A, B, C of an equilateral triangle of side $a$. The force between any two of them is

   $F=k\frac{q^2}{a^2},\qquad k=\frac{1}{4\pi\varepsilon_0}.$

   The fourth charge $3q$ is at the midpoint M of side AB.

   Step 1 - Forces from A and B.

   M lies on the line AB, at distance $a/2$ from each of A and B. Each exerts

   $F_{A}=F_{B}=k\frac{(q)(3q)}{(a/2)^2}=12\,k\frac{q^2}{a^2}=12F.$

   Because A and B are on opposite sides of M along the same straight line, these two repulsive forces point in exactly opposite directions. Being equal in magnitude, they cancel completely.

   Step 2 - Force from C.

   The distance from M to the opposite vertex C is the altitude of the triangle:

   $MC=\frac{\sqrt{3}}{2}a.$

   The force from C is …

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8. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Two point charges $+2\,\mu\text{C}$ and $+8\,\mu\text{C}$ are placed at a distance of $15\,\text{cm}$ apart in air. At a point on the line joining these two charges where the net electric field becomes zero, a third charge of $+5\,\mu\text{C}$ is placed. The net electrostatic force acting on $+5\,\mu\text{C}$ charge is (A) $16\,\text{N}$ (B) $4\,\text{N}$ (C) $8\,\text{N}$ (D) Zero

   ›Reveal solutionSolution

   The third charge is placed at the point where the fields from the two fixed charges cancel. Since the net electric field is zero there, the net force on any charge placed at that point is zero — regardless of the charge’s magnitude or sign. The answer is zero.

   The key idea is deceptively simple: force on a charge equals charge times the net electric field at its location. If the net field is zero, the force is zero. The problem gives you the location — the point where the two fixed charges produce zero net field — so you don’t even need to calculate where that point is. The third charge is placed exactly there, so the force on it is zero.

   Let’s walk through it carefully.

   1. What the problem tells us

      Two point charges, $+2\,\mu\text{C}$ and $+8\,\mu\text{C}$, are $15\,\text{cm}$ apart. Somewhere on the line joining them, the net electric field is zero. That point is where the fields from the two charges are equal in magnitude and opposite in direction.

      A third charge $+5\,\mu\text{C}$ is placed at that very point.

   2. The relation between field and force

      The electrostatic force on a charge $q$ in an electric field $\vec{E}$ is

   $\vec{F} = q \vec{E}.$

   This is a direct proportionality. If $\vec{E} = 0$ at a point, then for any charge placed there, $\vec{F} = 0$.

   3. Why the field is zero at that point

      Both charges are positive, so their fields point away from each charge. On the line between them, the fields are opposite in direction. There is exactly one point where their magnitudes match, cancelling each other. That point is given in the problem as the location where the third charge is placed.

   4. The third charge’s effect …

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9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.Four bodies of masses 8 kg, 2 kg, 4 kg and 2 kg are placed at the four corners A, B, C and D respectively of a square ABCD of diagonal 80 cm. Distance of centre of mass of the system from the corner A is (A) 30 cm (B) 40 cm (C) 60 cm (D) 20 cm

   ›Reveal solutionSolution

   Place the square in a coordinate system with A at the origin, use the diagonal to find the side length, then compute the centre of mass using the weighted average formula. The distance from A is $\boxed{30\ \text{cm}}$.

   The centre of mass of a system of particles is the point where the entire mass can be thought to be concentrated for translational motion. For discrete masses, it’s simply the mass-weighted average of their positions. The trick here is to choose a convenient coordinate system — since the square’s geometry is symmetric but the masses are not, we can’t just guess the answer; we must calculate.

   The diagonal of the square is given as 80 cm. For a square of side $a$, the diagonal is $a\sqrt{2}$. So $a\sqrt{2} = 80$, giving $a = \frac{80}{\sqrt{2}} = 40\sqrt{2}$ cm. That’s the side length.

   Now place the square with corner A at the origin $(0,0)$. Let’s assign coordinates:

   1. Set up coordinates: Let A be at $(0,0)$. Then B is at $(a,0) = (40\sqrt{2}, 0)$, C is at $(a,a) = (40\sqrt{2}, 40\sqrt{2})$, and D is at $(0,a) = (0, 40\sqrt{2})$. The masses are: $m_A = 8\ \text{kg}$, $m_B = 2\ \text{kg}$, $m_C = 4\ \text{kg}$, $m_D = 2\ \text{kg}$.

   2. Find the x-coordinate of the centre of mass:

   $X_{\text{cm}} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D}$

   The total mass is $8+2+4+2 = 16\ \text{kg}$.

   $X_{\text{cm}} = \frac{8(0) + 2(40\sqrt{2}) + 4(40\sqrt{2}) + 2(0)}{16} = \frac{0 + 80\sqrt{2} + 160\sqrt{2} + 0}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2}\ \text{cm}$

   3. Find the y-coordinate of the centre of mass:

   $Y_{\text{cm}} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{16}$

   $Y_{\text{cm}} = \frac{8(0) + 2(0) + 4(40\sqrt{2}) + 2(40\sqrt{2})}{16} = \frac{0 + 0 + 160\sqrt{2} + 80\sqrt{2}}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2}\ \text{cm}$

   So the centre of mass is at $(15\sqrt{2}, 15\sqrt{2})$ cm from A. …

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10. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Two positive point charges of $10\,\mu\text{C}$ and $12\,\mu\text{C}$ are kept in air with a separation of $12\,\text{cm}$. To make the distance between the charges as $4\,\text{cm}$, the work done is (A) $24\,\text{J}$ (B) $18\,\text{J}$ (C) $9\,\text{J}$ (D) $12\,\text{J}$

    ›Reveal solutionSolution

    The work done to change the separation between two point charges is equal to the change in their electrostatic potential energy. The work done is $\boxed{18\,\text{J}}$.

    The problem asks for the work done to change the distance between two positive point charges. This work is done by an external agent against the electrostatic repulsive force between the charges. When an external agent does work to move charges against the electric field, this work is stored as electrostatic potential energy in the system.

    Therefore, the work done to change the configuration of the charges is simply the difference between the final electrostatic potential energy and the initial electrostatic potential energy of the system.

    The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ in air (or vacuum) is given by:

    $U = \frac{k q_1 q_2}{r}$

    where $k$ is Coulomb's constant, $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$.

    The work done $W$ by an external agent to change the separation from an initial distance $r_1$ to a final distance $r_2$ is:

    $W = U_{\text{final}} - U_{\text{initial}}$

    $W = \frac{k q_1 q_2}{r_2} - \frac{k q_1 q_2}{r_1}$

    $W = k q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$

    Let's calculate the work done step-by-step.

    1. Identify the given values and convert units:

       • Charge $q_1 = 10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C} = 10^{-5}\,\text{C}$
       • Charge $q_2 = 12\,\mu\text{C} = 12 \times 10^{-6}\,\text{C}$
       • Initial separation $r_1 = 12\,\text{cm} = 0.12\,\text{m}$
       • Final separation $r_2 = 4\,\text{cm} = 0.04\,\text{m}$
       • Coulomb's constant $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$

    2. Calculate the initial electrostatic potential energy ($U_1$):

       Using the formula $U = \frac{k q_1 q_2}{r}$:

       $U_1 = \frac{(9 \times 10^9 \, \text{N m}^2/\text{C}^2) (10^{-5}\,\text{C}) (12 \times 10^{-6}\,\text{C})}{0.12\,\text{m}}$ …

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11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.Two point charges of magnitudes $-8\,\mu \mathrm{C}$ and $+32\,\mu \mathrm{C}$ are separated by a distance of $15\,\mathrm{cm}$ in air. The position of the point from $-8\,\mu \mathrm{C}$ charge at which the resultant electric field becomes zero is (A) $15\,\mathrm{cm}$ (B) $30\,\mathrm{cm}$ (C) $7.5\,\mathrm{cm}$ (D) $5\,\mathrm{cm}$

    ›Reveal solutionSolution

    The key idea is that the electric fields from two opposite charges cancel only on the side of the smaller charge, outside the segment joining them. Solving the distance ratio from Coulomb’s law gives the zero-field point 15 cm from the −8 µC charge, so the correct option is (A).

    We have two point charges:

    $q_1 = -8\,\mu\mathrm{C}$ and $q_2 = +32\,\mu\mathrm{C}$, separated by $d = 15\,\mathrm{cm}$.

    We want the point where the net electric field is zero.

    ---

    Concept and intuition

    Electric field vectors from a positive charge point away from it; from a negative charge, they point toward it.

    Here, the charges are opposite in sign. That means the fields can cancel only outside the segment joining them, on the side of the smaller magnitude charge. Why?

    • Between the charges, both fields point in the same direction (toward the negative, away from the positive), so they add, not cancel.
    • On the side of the larger charge, the field from the larger charge dominates; the smaller charge’s field cannot match it.
    • On the side of the smaller charge, the fields point in opposite directions, so cancellation is possible at some distance.

    Thus, the zero-field point lies on the line extending from the $-8\,\mu\mathrm{C}$ charge away from the $+32\,\mu\mathrm{C}$ charge.

    ---

    Step-by-step solution

    1. Set up coordinates

       Place $q_1 = -8\,\mu\mathrm{C}$ at $x = 0$ and $q_2 = +32\,\mu\mathrm{C}$ at $x = 15\,\mathrm{cm}$.

       Let the point where $\vec{E} = 0$ be at $x = -r$ (to the left of $q_1$), so its distance from $q_1$ is $r$ and from $q_2$ is $r + 15$.

    2. Write the condition for zero net field

       At that point, the magnitudes of the fields from each charge must be equal (since they point opposite directions):

    $\frac{k |q_1|}{r^2} = \frac{k |q_2|}{(r+15)^2}$

    Cancel $k$ and substitute magnitudes:

    $\frac{8}{r^2} = \frac{32}{(r+15)^2}$

    3. Solve the equation Divide both sides by 8: $$$$

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12. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.Four identical particles each of mass ‘m’ are kept at the four corners of a square of side ‘a’. If one of the particles is removed, the shift in the position of the centre of mass is (A) $\sqrt{2}\, a$ (B) $\dfrac{3a}{\sqrt{2}}$ (C) $\dfrac{a}{\sqrt{2}}$ (D) $\dfrac{a}{3\sqrt{2}}$

    ›Reveal solutionSolution

    The centre of mass of a system shifts by the vector from the original COM to the COM of the remaining particles. Removing one corner particle from a square shifts the COM by $\frac{a}{3\sqrt{2}}$ toward the opposite corner, so the correct option is (D).

    ---

    Concept and intuition

    The centre of mass (COM) of a system of particles is the weighted average of their positions. When you remove a particle, you are effectively subtracting its contribution from the total mass and from the total moment (mass × position). The new COM is the COM of the remaining three particles. The shift is simply the vector difference between the new COM and the original COM. Because the original four particles are symmetric, their COM is at the centre of the square. Removing one particle breaks that symmetry, and the COM moves away from the removed particle toward the opposite corner. The magnitude of that shift can be found by treating the removed particle as a negative mass at its location.

    ---

    Step-by-step solution

    1. Set up coordinates

       Place the square in the $xy$-plane with corners at

       $(0,0)$, $(a,0)$, $(a,a)$, $(0,a)$.

       All four particles have mass $m$, so total mass $M = 4m$.

    2. Original centre of mass

       By symmetry, the COM of the four equal masses is at the centre of the square:

    $\mathbf{R}_{\text{old}} = \left( \frac{a}{2}, \frac{a}{2} \right).$

    3. Remove one particle

       Suppose we remove the particle at $(0,0)$. The remaining three particles are at $(a,0)$, $(a,a)$, $(0,a)$, each of mass $m$. Their total mass is $3m$.

    4. New centre of mass

       Compute the COM of the three remaining particles:

    $\mathbf{R}_{\text{new}} = \frac{ m(a,0) + m(a,a) + m(0,a) }{3m} = \frac{ (a+a+0,\; 0+a+a) }{3} = \left( \frac{2a}{3}, \frac{2a}{3} \right).$

    5. Shift vector The shift in the position of the COM is: $$\Delta \mathbf{R} = \mathbf{R}_{\text{new}} - \mathbf{R}_{\text{old}} = \left( \frac{2a}{3} - \frac{a}{2},\; \frac{2a}{3} - \frac{a}{2} \right) …$$

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