Q.A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A parallel plate capacitor consists of two circular plates of radii 5 cm and 10 cm. The centres of the two plates are kept on the same straight line with a separation of 'd' and the plates are completely immersed in a liquid of dielectric constant 18. If the capacitance of the capacitor immersed in the liquid is $250~\text{pF}$, then the value of 'd' is (A) 5 mm (B) 10 mm (C) 2.5 mm (D) 7.5 mm

   ›Reveal solutionSolution

   The effective area is that of the smaller plate; solve $C=\dfrac{\kappa\varepsilon_0 A}{d}$ for $d$, giving $d\approx5~\text{mm}$ (A).

   Step 1 — Effective plate area.

   For unequal plates the overlapping (effective) area is set by the smaller plate, radius $5~\text{cm}$:

   $A=\pi r^2 = \pi(0.05)^2 = 7.854\times10^{-3}~\text{m}^2$

   Step 2 — Apply the capacitance formula.

   $C=\frac{\kappa\varepsilon_0 A}{d}\;\Rightarrow\; d=\frac{\kappa\varepsilon_0 A}{C}$ …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The spheres A, B and C have radii $R$, $R$ and $2R$ respectively. Initially A has a charge $-Q$, B is neutral and C is positively charged with a surface charge density equal to that of A. If the three spheres are placed in contact with each other and later separated, then the charges on the three spheres A, B and C respectively are (A) $\dfrac{5Q}{4}$, $\dfrac{5Q}{4}$, $\dfrac{5Q}{2}$ (B) $Q$, $Q$, $Q$ (C) $\dfrac{6Q}{5}$, $\dfrac{6Q}{5}$, $\dfrac{3Q}{5}$ (D) $\dfrac{3Q}{4}$, $\dfrac{3Q}{4}$, $\dfrac{3Q}{2}$

   ›Reveal solutionSolution

   Conductors in contact redistribute charge until their potentials are equal. Conserving the total charge $3Q$ and imposing $V_A=V_B=V_C$ gives $\tfrac{3Q}{4},\ \tfrac{3Q}{4},\ \tfrac{3Q}{2}$ — option (D).

   Step 1 — Initial charges

   • Sphere A: $q_A = -Q$.
   • Sphere B: neutral, $q_B = 0$.
   • Sphere C (radius $2R$): its surface charge density equals that of A in magnitude but is positive, so

   $q_C = \sigma \times 4\pi(2R)^2 = \frac{Q}{4\pi R^2}\times 16\pi R^2 = +4Q.$

   Total charge (conserved):

   $Q_{\text{tot}} = -Q + 0 + 4Q = 3Q.$

   Step 2 — Equal-potential condition

   For a sphere of radius $r$ carrying charge $q$, $V = \dfrac{kq}{r}$. On contact all potentials equalise:

   $\frac{kQ_A}{R} = \frac{kQ_B}{R} = \frac{kQ_C}{2R} \;\Longrightarrow\; Q_A = Q_B, \qquad Q_C = 2Q_A.$

   Step 3 — Apply charge conservation …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.When two electric charges are placed at a distance apart in air, the electrostatic force between them is 400 N. If $\dfrac{1}{3}$ of the space between the two charges is filled with a medium of dielectric constant 4, then the electrostatic force between them is (A) 600 N (B) 225 N (C) 1200 N (D) 300 N

   ›Reveal solutionSolution

   When a dielectric slab fills part of the space between two charges, it effectively increases the "electrical distance" between them, thereby reducing the electrostatic force. The new force is $\boxed{225 \text{ N}}$.

   Concept and Intuition: How Dielectrics Affect Electrostatic Force

   At the heart of this problem lies Coulomb's Law, which describes the electrostatic force between two point charges. In a vacuum (or approximately in air), this force is given by:

   $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$

   where $q_1$ and $q_2$ are the magnitudes of the charges, $d$ is the distance between them, and $\epsilon_0$ is the permittivity of free space.

   When a dielectric medium is introduced, it gets polarized by the electric field of the charges. This polarization creates an internal electric field that opposes the original field, effectively reducing the net electric field within the dielectric. If the entire space between the charges were filled with a dielectric of constant $K$, the force would simply become $F' = F/K$.

   However, this problem presents a more nuanced scenario: only a portion of the space is filled with the dielectric. In such cases, the electric field lines from one charge to the other pass through both the air and the dielectric. The presence of the dielectric "stretches" the effective electrical path length between the charges.

   The key insight for this specific type of problem (point charges with a dielectric slab inserted between them) is to use the concept of an effective distance. A dielectric slab of thickness $t$ and dielectric constant $K$ placed between two charges separated by a total distance $d$ makes the system behave as if the charges were separated by a larger "effective distance" in air. This effective distance, $d_{eff}$, accounts for the reduced field strength within the dielectric. Since the force is inversely proportional to the square of the distance, an increased effective distance leads to a reduced force, which is consistent with the nature of dielectrics.

   Step-by-Step Solution

   1. Initial Force in Air: Let the two charges be $q_1$ and $q_2$, and the initial distance between them be $d$. The electrostatic force between them in air is given by Coulomb's Law:

   $F_{\text{air}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$

   We are given that $F_{\text{air}} = 400 \text{ N}$.
   

   2. Introducing the Dielectric Slab:

   A dielectric medium of dielectric constant $K=4$ is introduced. It fills $\frac{1}{3}$ of the space between the charges. This means the thickness of the dielectric slab, $t$, is $\frac{d}{3}$.

   The remaining space, which is air, has a thickness of $d - t = d - \frac{d}{3} = \frac{2d}{3}$.

   3. Calculating the Effective Distance: When a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between two charges separated by a total distance $d$, the system behaves as if the charges are separated by an "effective distance" $d_{eff}$ in air. This effective distance is given by the formula:

   $d_{eff} = (d - t) + t\sqrt{K}$

   This formula accounts for the way the electric field lines are modified by the dielectric. The term $(d-t)$ represents the air gap, and $t\sqrt{K}$ represents the equivalent air thickness for the dielectric region. Since $K>1$, $t\sqrt{K} > t$, meaning the dielectric effectively "stretches" the distance.
   
   Let's substitute the given values:
   

   $d_{eff} = \left(d - \frac{d}{3}\right) + \left(\frac{d}{3}\right)\sqrt{4}$

   $d_{eff} = \frac{2d}{3} + \left(\frac{d}{3}\right) \times 2$

   $d_{eff} = \frac{2d}{3} + \frac{2d}{3}$

   $$d_{eff} = \frac{4d}{3}$$ …
   

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A parallel plate capacitor of capacitance $5~\mu\mathrm{F}$ is connected to an ac source. If the relation between the potential difference $V$ (in volt) across the plates of the capacitor and the time $t$ (in second) is $V = 7\cos(100\pi t)$, then the displacement current between the plates of the capacitor at a time of $5~\mathrm{ms}$ is (A) $5~\mathrm{mA}$ (B) $9~\mathrm{mA}$ (C) $7~\mathrm{mA}$ (D) $11~\mathrm{mA}$

   ›Reveal solutionSolution

   Displacement current equals the conduction current in the circuit. For a capacitor, $I = C\frac{dV}{dt}$. Differentiate $V = 7\cos(100\pi t)$, evaluate at $t = 5~\mathrm{ms}$, and multiply by $C = 5~\mu\mathrm{F}$ to get $11~\mathrm{mA}$.

   The key idea is that displacement current is not some separate mysterious quantity — it is exactly the current that flows in the wire leading to the capacitor. Maxwell showed that the displacement current between the plates equals the conduction current in the circuit. So to find the displacement current, we simply find the current through the capacitor.

   For a capacitor, the current is related to the rate of change of voltage: $I = C \frac{dV}{dt}$. This is a fundamental relation that comes from $Q = CV$ and $I = dQ/dt$. No need to think about electric flux or Maxwell's equations for this problem — just differentiate the given voltage and plug in.

   1. Write down the given data

      Capacitance: $C = 5~\mu\mathrm{F} = 5 \times 10^{-6}~\mathrm{F}$

      Voltage: $V(t) = 7\cos(100\pi t)$ volts

      Time of interest: $t = 5~\mathrm{ms} = 5 \times 10^{-3}~\mathrm{s}$

   2. Find the current through the capacitor

      $I(t) = C \frac{dV}{dt} = C \cdot \frac{d}{dt}[7\cos(100\pi t)]$

      The derivative of $\cos(100\pi t)$ is $-100\pi \sin(100\pi t)$. So:

      $I(t) = 5 \times 10^{-6} \times 7 \times (-100\pi) \sin(100\pi t)$

      $I(t) = -35 \times 10^{-4} \pi \sin(100\pi t)$

      $I(t) = -3.5 \times 10^{-3} \pi \sin(100\pi t)$ amperes

   3. Evaluate at $t = 5~\mathrm{ms}$

      First compute the argument: $100\pi t = 100\pi \times 5 \times 10^{-3} = 0.5\pi$ radians

      So $\sin(0.5\pi) = \sin(\pi/2) = 1$

      Therefore: $I(5~\mathrm{ms}) = -3.5 \times 10^{-3} \pi \times 1 = -3.5\pi \times 10^{-3}~\mathrm{A}$

   4. Compute the numerical value …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.A thin spherical shell of radius $R$ and surface charge density $\sigma$ is placed in a cube of side $5R$ with their centers coinciding. The electric flux through one face of the cube is ($\epsilon_0$ = Permittivity of free space) (A) $\dfrac{2\pi R^2 \sigma}{3\epsilon_0}$ (B) $\dfrac{\pi R^2 \sigma}{3\epsilon_0}$ (C) $\dfrac{\sigma}{6\epsilon_0}$ (D) $\dfrac{\sigma}{4\pi \epsilon_0 R^2}$

   ›Reveal solutionSolution

   The key idea is that the total flux through the cube equals the charge enclosed divided by $\epsilon_0$, and by symmetry each of the six faces gets an equal share. The correct option is (A).

   Concept and intuition

   The problem asks for the electric flux through one face of a cube that contains a thin spherical shell of charge at its center. The shell is entirely inside the cube (since its radius $R$ is less than half the cube’s side $5R$). By Gauss’s law, the total flux through the entire cube is $Q_{\text{enc}}/\epsilon_0$, where $Q_{\text{enc}}$ is the total charge on the shell. Because the cube is symmetric with respect to the shell’s center, the flux is evenly distributed among the six faces. So we just need the total charge on the shell, divide by $6\epsilon_0$, and we’re done.

   Step-by-step solution

   1. Find the total charge on the spherical shell. The surface area of a sphere of radius $R$ is $4\pi R^2$. With uniform surface charge density $\sigma$, the total charge is

   $Q = \sigma \cdot (4\pi R^2) = 4\pi R^2 \sigma.$

   2. Apply Gauss’s law to the cube. The cube encloses the entire shell, so the net charge inside is $Q$. The total electric flux through the closed surface of the cube is

   $\Phi_{\text{total}} = \frac{Q}{\epsilon_0} = \frac{4\pi R^2 \sigma}{\epsilon_0}.$

   3. Use symmetry to find the flux through one face. The cube is centered on the shell’s center, so by symmetry each of the six faces has the same flux. Therefore, $$$$

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the magnetic susceptibility of a substance is 0.6, then the ratio of permeability of the substance and permeability of free space is (A) $6 : 5$ (B) $7 : 4$ (C) $8 : 5$ (D) $3 : 5$

   ›Reveal solutionSolution

   The magnetic susceptibility $\chi_m = 0.6$ relates to relative permeability $\mu_r$ by $\mu_r = 1 + \chi_m$, so $\mu_r = 1.6 = \frac{8}{5}$, giving the ratio $\mu : \mu_0 = 8 : 5$, which corresponds to option (C).

   The key concept here is the relationship between magnetic susceptibility and relative permeability. Susceptibility ($\chi_m$) tells us how much a material becomes magnetized in response to an external magnetic field. Permeability ($\mu$) measures how well the material supports the formation of a magnetic field inside it. The two are linked by a simple formula: $\mu_r = 1 + \chi_m$, where $\mu_r = \mu / \mu_0$ is the relative permeability (the ratio we want).

   Why this works:

   Magnetic susceptibility accounts for the material's own magnetization adding to (or subtracting from) the external field. For a linear, isotropic material, the total magnetic field inside is the sum of the applied field and the material's response. This sum is captured by the permeability. So the ratio $\mu / \mu_0$ is just $1$ plus the susceptibility.

   Now, let's work it through:

   1. Write the given data:

      Magnetic susceptibility, $\chi_m = 0.6$.

   2. Recall the defining relation:

      For any magnetic material (except superconductors in certain states),

   $\mu_r = 1 + \chi_m$

   where $\mu_r = \frac{\mu}{\mu_0}$ is the relative permeability.

   3. Substitute the value:

   $\mu_r = 1 + 0.6 = 1.6$

   4. Express as a ratio: $1.6 = \frac{16}{10} = \frac{8}{5}$. Therefore, $$$$

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.When the right gap of a meter bridge consists of two equal resistors in series, the balancing point is at 50 cm. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap, the balancing point is at (A) 60 cm (B) 33.3 cm (C) 25 cm (D) 40 cm

   ›Reveal solutionSolution

   The key is to track how the effective resistance in each gap changes when a resistor is moved from series to parallel. The final balance point is 40 cm, so the correct option is (D).

   ---

   Concept and Intuition

   A meter bridge is a Wheatstone bridge with a uniform wire of length 100 cm. The balance condition is:

   $\frac{R_{\text{left}}}{R_{\text{right}}} = \frac{l}{100 - l}$

   where $l$ is the distance from the left end to the balance point.

   When resistors are rearranged, the effective resistance in each gap changes. The trick is to express everything in terms of a single unknown resistance, then solve for the new balance length.

   ---

   Step-by-step solution

   1. Initial setup Let each of the two equal resistors in the right gap have resistance $R$. Initially, they are in series, so the right gap resistance is $R_{\text{right}} = R + R = 2R$. The left gap has some unknown resistor, call it $X$. Balance at 50 cm means:

   $\frac{X}{2R} = \frac{50}{50} = 1 \quad \Rightarrow \quad X = 2R.$

   So the left gap resistor is $2R$.

   2. New arrangement One resistor from the right gap is removed. That leaves a single resistor $R$ in the right gap. The removed resistor is connected in parallel with the left gap resistor $2R$. The new left gap resistance is the parallel combination:

   $R_{\text{left, new}} = \frac{(2R) \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R.$

   3. New balance condition

      The right gap now has just $R$. Let the new balance length be $l$ cm. Then:

      $\frac{R_{\text{left, new}}}{R_{\text{right, new}}} = \frac{l}{100 - l} \quad \Rightarrow \quad \frac{\frac{2}{3}R}{R} = \frac{l}{100 - l}.$ …

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The ratio of radii of gyration of a thin circular ring and a circular disc of same radius about a tangential axis in their own planes is $\sqrt{12} : \sqrt{K}$. The value of K is (A) 10 (B) 24 (C) 5 (D) 12

   ›Reveal solutionSolution

   The ratio of radii of gyration for a ring and a disc about a tangential axis in their plane is given as $\sqrt{12}:\sqrt{K}$, and by applying the parallel axis theorem to their moments of inertia about a diameter, we find $K = 5$.

   Concept & Intuition

   The radius of gyration $k$ is defined by $I = M k^2$, where $I$ is the moment of inertia about the axis. Here, the axis is tangential — a line tangent to the ring or disc and lying in its plane. That axis is parallel to a diameter of the object. So we first find the moment of inertia about a diameter (easy, from standard results), then use the parallel axis theorem to shift to the tangential axis. The ratio of the two $k$ values is given, and we solve for $K$.

   Step-by-step solution

   1. Moment of inertia about a diameter For a thin circular ring of mass $M$ and radius $R$, the moment of inertia about any diameter is

   $I_{\text{ring, diam}} = \frac{1}{2} M R^2.$

   For a solid circular disc of same mass and radius, about a diameter:

   $I_{\text{disc, diam}} = \frac{1}{4} M R^2.$

   2. Shift to a tangential axis in the plane A tangential axis in the plane is parallel to a diameter and lies at a distance $R$ from the center (since the tangent touches the circle at the rim). By the parallel axis theorem:

   $I_{\text{tangent}} = I_{\text{diam}} + M R^2.$

   So for the ring:

   $I_{\text{ring, tan}} = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2.$

   For the disc:

   $I_{\text{disc, tan}} = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2.$

   3. Radii of gyration Since $I = M k^2$, we have

   $k_{\text{ring}} = \sqrt{\frac{I_{\text{ring, tan}}}{M}} = \sqrt{\frac{3}{2}} R,$

   $k_{\text{disc}} = \sqrt{\frac{I_{\text{disc, tan}}}{M}} = \sqrt{\frac{5}{4}} R = \frac{\sqrt{5}}{2} R.$

   4. Form the given ratio The problem states the ratio is $\sqrt{12} : \sqrt{K}$. That means

   $\frac{k_{\text{ring}}}{k_{\text{disc}}} = \frac{\sqrt{12}}{\sqrt{K}}.$

   Substitute the expressions:

   $\frac{\sqrt{\frac{3}{2}} R}{\frac{\sqrt{5}}{2} R} = \frac{\sqrt{12}}{\sqrt{K}}.$ …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The ratio of radii of gyration of a thin circular ring and a circular disc of same radius about a tangential axis in their own planes is $\sqrt{12} : \sqrt{K}$. The value of K is (A) 10 (B) 5 (C) 12 (D) 24

   ›Reveal solutionSolution

   The key idea is to apply the parallel axis theorem to find the moment of inertia about a tangential axis in the plane for both a ring and a disc, then compare their radii of gyration to solve for $K$. The result is $K = 5$.

   Concept & Intuition

   The radius of gyration $k$ is defined by $I = M k^2$, where $I$ is the moment of inertia about a given axis. For a thin circular ring and a disc of the same radius $R$, we need the moment of inertia about a tangential axis lying in their own plane — that is, an axis that just touches the object at one point and lies flat in the plane of the object.

   The trick: we know the moment of inertia about a diameter (an axis through the center in the plane). Then we use the parallel axis theorem to shift that axis to the edge (tangent). This avoids messy integration.

   ---

   Step-by-step solution

   1. Moment of inertia of a thin circular ring about a diameter For a ring of mass $M$ and radius $R$, the moment of inertia about any diameter is

   $I_{\text{ring, diam}} = \frac{1}{2} M R^2.$

   (This is a standard result: the ring has all mass at distance $R$ from the center, but for a diameter, the perpendicular distance varies; the average gives $\frac12 M R^2$.)

   2. Shift to a tangential axis in the plane (ring) A tangential axis in the plane is parallel to a diameter but shifted by distance $R$ (from center to edge). By the parallel axis theorem:

   $I_{\text{ring, tan}} = I_{\text{ring, diam}} + M R^2 = \frac12 M R^2 + M R^2 = \frac32 M R^2.$

   Hence the radius of gyration for the ring about this axis is

   $k_{\text{ring}} = \sqrt{\frac{I_{\text{ring, tan}}}{M}} = \sqrt{\frac32}\, R.$

   3. Moment of inertia of a circular disc about a diameter For a solid disc of mass $M$ and radius $R$, the moment of inertia about a diameter is

   $I_{\text{disc, diam}} = \frac14 M R^2.$

   (Standard result: for a disc, $I$ about a diameter is half of $I$ about the perpendicular axis through center, which is $\frac12 M R^2$.)

   4. Shift to a tangential axis in the plane (disc) Again, the tangential axis is parallel to a diameter and at distance $R$ from the center. Using the parallel axis theorem:

   $I_{\text{disc, tan}} = I_{\text{disc, diam}} + M R^2 = \frac14 M R^2 + M R^2 = \frac54 M R^2.$

   So the radius of gyration for the disc is

   $k_{\text{disc}} = \sqrt{\frac54}\, R.$

   5. Set up the given ratio The problem states:

   $\frac{k_{\text{ring}}}{k_{\text{disc}}} = \frac{\sqrt{12}}{\sqrt{K}}.$

   Substitute the expressions:

   $$$$

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.The electrostatic force between two charges kept in air is F. If 30% of the space between the charges is filled with a medium, then the electrostatic force between the charges becomes $\dfrac{F}{2.56}$. The dielectric constant of the medium is (A) 8 (B) 3 (C) 9 (D) 4

    ›Reveal solutionSolution

    The key idea is that the effective dielectric constant for a partially filled space is the weighted harmonic mean of the permittivities of air and the medium, weighted by the fraction of space each occupies. Solving gives the dielectric constant as 4, so the correct option is (D).

    Concept & Intuition

    When two charges are in air, the force is given by Coulomb’s law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$. If we fill part of the space between them with a dielectric, the electric field lines pass partly through air and partly through the medium. The effective permittivity is not a simple average — because the field lines are in series (they go through one region then the other), the reciprocals of the permittivities add in proportion to the path lengths. This is analogous to capacitors in series: the total capacitance is the harmonic mean of the individual capacitances. Here, the force is inversely proportional to the effective permittivity, so we set up a weighted harmonic mean.

    Step-by-step solution

    1. Define the fractions.

       Let the total distance between the charges be $r$. The problem says 30% of the space is filled with the medium. That means the medium occupies a length $0.3r$ and air occupies the remaining $0.7r$. (We assume the medium is placed as a contiguous slab, which is the standard interpretation.)

    2. Force in air alone.

       In air (dielectric constant $K_{\text{air}} = 1$), the force is

    $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}.$

    3. Force with partial dielectric. When the space is partially filled, the effective permittivity $\varepsilon_{\text{eff}}$ is given by the series combination:

    $\frac{r}{\varepsilon_{\text{eff}}} = \frac{0.7r}{\varepsilon_0} + \frac{0.3r}{K\varepsilon_0},$

    where $K$ is the dielectric constant of the medium. Cancel $r$ and $\varepsilon_0$:

    $\frac{1}{\varepsilon_{\text{eff}}} = \frac{0.7}{\varepsilon_0} + \frac{0.3}{K\varepsilon_0} = \frac{1}{\varepsilon_0}\left(0.7 + \frac{0.3}{K}\right).$

    Hence

    $\varepsilon_{\text{eff}} = \frac{\varepsilon_0}{0.7 + \frac{0.3}{K}}.$

    4. New force expression. The force becomes

    $F' = \frac{1}{4\pi\varepsilon_{\text{eff}}} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \cdot \frac{1}{\frac{1}{\varepsilon_0}\varepsilon_{\text{eff}}?}$

    Actually, simpler:

    $F' = \frac{1}{4\pi\varepsilon_{\text{eff}}} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \cdot \frac{\varepsilon_0}{\varepsilon_{\text{eff}}} = F \cdot \frac{\varepsilon_0}{\varepsilon_{\text{eff}}}.$

    Substitute $\varepsilon_{\text{eff}}$:

    $\frac{\varepsilon_0}{\varepsilon_{\text{eff}}} = 0.7 + \frac{0.3}{K}.$

    So

    $F' = F \left(0.7 + \frac{0.3}{K}\right).$

    5. Use the given ratio. The problem states $F' = \frac{F}{2.56}$. Therefore:

    $0.7 + \frac{0.3}{K} = \frac{1}{2.56}.$

    Compute $\frac{1}{2.56} = 0.390625$. So:

    $0.7 + \frac{0.3}{K} = 0.390625.$

    This gives

    $\frac{0.3}{K} = 0.390625 - 0.7 = -0.309375.$

    That’s negative — impossible. Wait: this signals a classic pitfall.

    Watch out

    The force decreases when a dielectric is inserted, so $F' < F$. Here $F' = F/2.56$ is indeed smaller, but our equation gave a negative term because we assumed the medium replaces air. Actually, the 30% filled means the medium occupies 30% of the line between charges, but the force formula uses the inverse of the effective permittivity. The correct relation is:

    $\frac{1}{\varepsilon_{\text{eff}}} = \frac{0.7}{\varepsilon_0} + \frac{0.3}{K\varepsilon_0}$

    leads to $\varepsilon_{\text{eff}} > \varepsilon_0$ only if $K>1$, so $F' < F$. Our algebra is correct; the negative result means the given ratio $1/2.56$ is too small for $K>1$? Let’s re-check: $0.7 + 0.3/K$ must be less than 1. For $K=4$, $0.7+0.075=0.775$, which is > 0.3906. So something is off: perhaps the 30% refers to the volume fraction, not the linear fraction? But in a parallel-plate-like geometry, the fraction of the distance is what matters. Let’s re-interpret: The problem likely means that 30% of the space between the charges is filled, so the medium occupies 30% of the total distance. Then the effective dielectric constant is the weighted harmonic mean of the permittivities, but the force is inversely proportional to the effective permittivity. Wait — we might have inverted the relation.

    6. Correct the series formula. For two capacitors in series (air gap and dielectric gap), the total capacitance is

    $\frac{1}{C_{\text{total}}} = \frac{1}{C_{\text{air}}} + \frac{1}{C_{\text{dielectric}}}.$

    Capacitance is proportional to $\varepsilon / d$, so

    $\frac{d_{\text{total}}}{\varepsilon_{\text{eff}}} = \frac{d_{\text{air}}}{\varepsilon_0} + \frac{d_{\text{dielectric}}}{K\varepsilon_0}.$

    With $d_{\text{total}} = r$, $d_{\text{air}} = 0.7r$, $d_{\text{dielectric}} = 0.3r$, we get

    $\frac{r}{\varepsilon_{\text{eff}}} = \frac{0.7r}{\varepsilon_0} + \frac{0.3r}{K\varepsilon_0} \quad \Rightarrow \quad \frac{1}{\varepsilon_{\text{eff}}} = \frac{1}{\varepsilon_0}\left(0.7 + \frac{0.3}{K}\right).$

    That is correct. Then the force $F' \propto 1/\varepsilon_{\text{eff}}$, so

    $F' = F \cdot \frac{\varepsilon_0}{\varepsilon_{\text{eff}}} = F\left(0.7 + \frac{0.3}{K}\right).$

    Given $F' = F/2.56$, we have

    $0.7 + \frac{0.3}{K} = \frac{1}{2.56} \approx 0.3906.$

    This implies $0.3/K = -0.3094$, impossible. So the only logical conclusion is that the 30% refers to the volume of the space, but in a 3D sense? Actually, for point charges, the field lines are not confined to a cylinder; the concept of “filling 30% of the space” is ambiguous. The standard textbook trick: when a dielectric slab of thickness $t$ is inserted between two charges, the effective distance changes. The correct formula is:

    $F' = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{(r - t + t\sqrt{K})^2} \quad \text{? No, that's for a slab in a capacitor.}$

    For point charges, the force is not simply given by a series capacitance model because the field is not uniform. However, many problems approximate it as such. Let’s instead solve the equation as given, assuming the series model is correct but the fraction is of the distance. The negative result suggests the given ratio $1/2.56$ is actually $1/2.56 = 0.390625$, and if we set $0.7 + 0.3/K = 0.3906$, then $K$ would be negative. So perhaps the 30% is the fraction of the medium in the total effective permittivity? Let’s try the other common model: effective dielectric constant is the weighted arithmetic mean? That would give $K_{\text{eff}} = 0.7(1) + 0.3K$, and force $F' = F/K_{\text{eff}}$. Then

    $\frac{1}{0.7 + 0.3K} = \frac{1}{2.56} \quad \Rightarrow \quad 0.7 + 0.3K = 2.56 \quad \Rightarrow \quad 0.3K = 1.86 \quad \Rightarrow \quad K = 6.2,$

    not an option. So that’s not it.

    7. Re-examine the problem statement. It says “30% of the space between the charges is filled with a medium”. In many textbooks, this means that the medium occupies 30% of the volume of the space, but the field lines are assumed to be parallel (like in a parallel-plate capacitor). Then the effective dielectric constant is the volume-weighted harmonic mean of the permittivities? Actually, for parallel plates with two dielectrics in series, the effective dielectric constant is the harmonic mean of the thickness fractions. For parallel (side-by-side) dielectrics, it’s the arithmetic mean. Here, since the medium is “between” the charges, it’s series. So the series formula is correct. The only way to get a positive $K$ is if the fraction is of the medium in the total effective permittivity differently. Let’s solve the equation properly:

    $0.7 + \frac{0.3}{K} = \frac{1}{2.56} = 0.390625$

    $\frac{0.3}{K} = -0.309375 \quad \Rightarrow \quad K = -\frac{0.3}{0.309375} \approx -0.97.$

    Impossible. So the given ratio must be $F' = F/2.56$ meaning the force is reduced by a factor of 2.56, so $F' = F/2.56$ implies $0.7 + 0.3/K = 1/2.56$? That gives a negative. Wait — maybe the force becomes $F/2.56$ means it is larger? No, dielectric reduces force. So $F' < F$, so the factor should be less than 1. $1/2.56 \approx 0.39$ is less than 1, so that’s fine. But our equation gives a negative. The only resolution: the 30% is the fraction of the medium in the total distance, but the formula should be

    $\frac{1}{\varepsilon_{\text{eff}}} = \frac{0.3}{\varepsilon_0 K} + \frac{0.7}{\varepsilon_0} \quad \text{(same)}.$

    Let’s instead assume the medium fills 30% of the volume but the field lines go through both in parallel? That would give arithmetic mean: $\varepsilon_{\text{eff}} = 0.3K\varepsilon_0 + 0.7\varepsilon_0$, then $F' = F / (0.3K + 0.7)$. Set equal to $F/2.56$:

    $0.3K + 0.7 = 2.56 \quad \Rightarrow \quad 0.3K = 1.86 \quad \Rightarrow \quad K = 6.2,$

    not an option. So that’s not it. …

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.The electrostatic force between two charges kept in air is F. If 30% of the space between the charges is filled with a medium, then the electrostatic force between the charges becomes $\dfrac{F}{2.56}$. The dielectric constant of the medium is (A) 8 (B) 9 (C) 3 (D) 4

    ›Reveal solutionSolution

    A dielectric slab of thickness $t=0.3d$ makes the effective separation $d_{\text{eff}}=0.7d+0.3d\sqrt{K}$; from $F'=F/2.56$ we get $\sqrt{K}=3$, so $K=9$.

    Let the charges be a distance $d$ apart in air, giving force

    $F=\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{d^{2}}.$

    A slab of dielectric constant $K$ and thickness $t=0.30\,d$ fills $30\%$ of the gap; the remaining $0.70\,d$ is air. A dielectric slab of thickness $t$ increases the effective separation to

    $d_{\text{eff}}=(d-t)+t\sqrt{K}=0.7d+0.3d\sqrt{K}.$

    The new force is

    $F'=\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{d_{\text{eff}}^{2}}=F\left(\frac{d}{d_{\text{eff}}}\right)^{2}.$ …

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12. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.In a meter bridge, when an unknown resistance ‘R’ is connected in the left gap, the null point is obtained at 25 cm from the left end of the wire. If the resistance in the left gap is increased by 100%, the distance of the null point from the left end of the wire increases by (A) 50% (B) 40% (C) 60% (D) 80%

    ›Reveal solutionSolution

    The meter bridge null condition gives a ratio of resistances equal to the ratio of wire lengths. Increasing the left resistance by 100% shifts the null point from 25 cm to 40 cm, a 60% increase in distance.

    The meter bridge works on the principle of a Wheatstone bridge. When the galvanometer shows zero deflection, the ratio of the two resistances in the left and right gaps equals the ratio of the lengths of the wire on either side of the null point. This is because the wire has uniform resistance per unit length, so the resistance of a segment is proportional to its length.

    Let the unknown resistance in the left gap be $R$, and let the known resistance in the right gap be $S$ (which remains constant throughout). The null point is at 25 cm from the left end, so the left segment length is $L_1 = 25$ cm and the right segment length is $L_2 = 100 - 25 = 75$ cm.

    1. First condition: At the initial null point, the bridge is balanced:

    $\frac{R}{S} = \frac{L_1}{L_2} = \frac{25}{75} = \frac{1}{3}$

    So $S = 3R$.

    2. Second condition: The resistance in the left gap is increased by 100%, meaning it becomes $R' = R + 100\% \text{ of } R = 2R$. The right gap resistance $S$ is unchanged. Let the new null point be at distance $x$ cm from the left end. Then the left segment length is $x$ cm and the right segment length is $(100 - x)$ cm. The new balance condition is:

    $\frac{2R}{S} = \frac{x}{100 - x}$

    3. Substitute $S = 3R$ into the second equation:

    $\frac{2R}{3R} = \frac{x}{100 - x} \quad \Rightarrow \quad \frac{2}{3} = \frac{x}{100 - x}$

    4. Solve for $x$: $2(100 - x) = 3x \quad \Rightarrow \quad 200 - 2x = 3x \quad \Rightarrow \quad 200 = 5x \quad \Rightarrow \quad x = 40 \text{ cm}$ …

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