A container has a base of 50 cm × 5 cm and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm × 50 cm. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250 cm3s−1. What is the value of the capacitance of the container after 10 seconds? [Given: Permittivity of free space 𝜖0 = 9 × 10−12 C2N−1m−2, the effects of the non-conducting walls on the capacitance are negligible]
A container has a base of 50 cm × 5 cm and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm × 50 cm. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250 cm3s−1. What is the value of the capacitance of the container after 10 seconds? [Given: Permittivity of free space 𝜖0 = 9 × 10−12 C2N−1m−2, the effects of the non-conducting walls on the capacitance are negligible]
27 pF
63 pF
81 pF
135 pF
The Correct Option is B
Approach Solution - 1
To determine the capacitance of the container after 10 seconds, we need to calculate the change in height of the liquid using its filling rate and then determine how this affects the capacitance.
Given:
- Base dimensions: 50 cm × 5 cm
- Height of container: 50 cm
- Filling rate: 250 cm3/s
- Dielectric constant (κ): 3
- Permittivity of free space (𝜖0): 9 × 10−12 C²/N·m²
1. Calculate the volume of liquid filled in 10 seconds:
Volumetric flow rate = 250 cm3/s
Total volume of liquid filled in 10 seconds = 250 cm³/s × 10 s = 2500 cm³
2. Determine the height of the liquid column:
Base area = 50 cm × 5 cm = 250 cm²
Height of the liquid (h) after 10 seconds = Volume / Base area = 2500 cm³ / 250 cm² = 10 cm
3. Calculate the capacitance:
The capacitance of a parallel plate capacitor with a dielectric is given by:
C = (κ × 𝜖0 × A) / d
Where:
- κ = Dielectric constant = 3
- 𝜖0 = 9 × 10−12 C²/N·m²
- A = Area of one plate = 50 cm × 50 cm = 2500 cm² = 0.25 m²
- d = Distance between plates (height of liquid) = 10 cm = 0.1 m
C = (3 × 9 × 10−12 × 0.25) / 0.1 = 67.5 × 10−12 F = 67.5 pF
The closest option to this calculated value is 63 pF.
Approach Solution -2
\(h = \frac {250*10}{50*5} = 10 cm \)
\[C1 = \frac{(0.40 * 0.50) * 9 * 10}{5 * 10}\]= 0.36 * 10-10 F
\(C2 = \frac {3*0.10*0.5*9*10-12}{5*10-2}\)
C2 = 0.27 * 10-10 F
C = C1 + C2
= 63pF
Top Questions on Electrostatics
Three parallel plate capacitors each with area \(A\) and separation \(d\) are filled with two dielectric (\(k_1\) and \(k_2\)) in the following fashion. (\(k_1>k_2\)) Which of the following is true?
- JEE Main - 2026
- Physics
- Electrostatics
- In the following configuration of charges. Find the net dipole moment of the system :
- JEE Main - 2026
- Physics
- Electrostatics
- Five positive charges each having charge \(q\) are placed at the vertices of a regular pentagon as shown in the figure. The electric potential \(V\) and the electric field \(\vec{E}\) at the center \(O\) of the pentagon due to these five positive charges are
- JEE Main - 2026
- Physics
- Electrostatics
A point charge \(q = 1\,\mu\text{C}\) is located at a distance \(2\,\text{cm}\) from one end of a thin insulating wire of length \(10\,\text{cm}\) having a charge \(Q = 24\,\mu\text{C}\), distributed uniformly along its length, as shown in the figure. Force between \(q\) and wire is ________ N.
- JEE Main - 2026
- Physics
- Electrostatics
- The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^{3} + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $a \times \pi \varepsilon_{0}$. The value of $a$ is ___.
(Permittivity of vacuum is $\varepsilon_{0}$)
- JEE Main - 2026
- Physics
- Electrostatics
Questions Asked in JEE Advanced exam
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16- JEE Advanced - 2025
- Biomolecules
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Conic sections
- Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$
- JEE Advanced - 2025
- Adsorption
- At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$
- JEE Advanced - 2025
- Colligative Properties
Concepts Used:
Electrostatic Potential
The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as “The amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.”
SI Unit of Electrostatic Potential:
SI unit of electrostatic potential - volt
Other units - statvolt
Symbol of electrostatic potential - V or φ
Dimensional formula - ML2T3I-1
Electric Potential Formula:
The electric potential energy of the system is given by the following formula:
U = 1/(4πεº) × [q1q2/d]
Where q1 and q2 are the two charges that are separated by the distance d.