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
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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.