The nature of the capacitance of an electrostatic capacitor depends on:
A) Shape
B) Size
C) Thickness of plate
D) Area
Answer
253.2k+ views
Hint: Capacitance of a capacitor can be calculated if we know the area of the plates, the distance between the plates.
Formula used:
The capacitance $C$ of a capacitor with area $A$ and distance $d$ between the plates is given by, $C = \dfrac{{{\varepsilon _0}A}}{d}$ where ${\varepsilon _0}$ is the permittivity of vacuum.
Complete step by step answer:
Define capacitance.
A capacitor is a system of two conductors with charges $Q$ and $ - Q$ separated by an insulator. There exists a potential difference $V$ between these two conductors. The total charge of the capacitor is zero.
Now, if the charge on the capacitor is doubled the electric field in the region between the two conductors will also double.
i.e., $Q \propto V$ or $C = \dfrac{Q}{V}$ .
Here, the constant of proportionality $C$ is called capacitance of a capacitor. It refers to the ability of the capacitor to store electric charge between its plates.
Obtain an equation for the capacitance.
For a uniform electric field, the potential $V$can be expressed as $V = Ed$ where $E$ is the electric field and $d$ is the distance between the plates.
Also, $E = \dfrac{Q}{{{\varepsilon _0}A}}$ .
Substituting for $E$ in $V = Ed$ we get,
$V = \dfrac{Q}{{{\varepsilon _0}A}}d$ .
We also have $V = \dfrac{Q}{C}$ . So, we equate the two equations to get, $\dfrac{Q}{C} = \dfrac{Q}{{{\varepsilon _0}A}}d$ .
Simplifying, we get $C = \dfrac{{{\varepsilon _0}A}}{d}$ .
This suggests that the capacitance increases if the area of the plates increases keeping the distance between the plates as a constant or the capacitance increases if there is a decrease in the separation of the plates.
A change in the size or shape of the capacitor plates is reflected as a change in the area of the plates.
Therefore, the correct options are A, B and D.
Note: Capacitance depends on the area of the plates, the separation between the plates and the nature of the dielectric material placed between the plates. The thickness of the plates do not contribute to the area of the plates and hence has no effect on the capacitance of the plate.
Formula used:
The capacitance $C$ of a capacitor with area $A$ and distance $d$ between the plates is given by, $C = \dfrac{{{\varepsilon _0}A}}{d}$ where ${\varepsilon _0}$ is the permittivity of vacuum.
Complete step by step answer:
Define capacitance.
A capacitor is a system of two conductors with charges $Q$ and $ - Q$ separated by an insulator. There exists a potential difference $V$ between these two conductors. The total charge of the capacitor is zero.
Now, if the charge on the capacitor is doubled the electric field in the region between the two conductors will also double.
i.e., $Q \propto V$ or $C = \dfrac{Q}{V}$ .
Here, the constant of proportionality $C$ is called capacitance of a capacitor. It refers to the ability of the capacitor to store electric charge between its plates.
Obtain an equation for the capacitance.
For a uniform electric field, the potential $V$can be expressed as $V = Ed$ where $E$ is the electric field and $d$ is the distance between the plates.
Also, $E = \dfrac{Q}{{{\varepsilon _0}A}}$ .
Substituting for $E$ in $V = Ed$ we get,
$V = \dfrac{Q}{{{\varepsilon _0}A}}d$ .
We also have $V = \dfrac{Q}{C}$ . So, we equate the two equations to get, $\dfrac{Q}{C} = \dfrac{Q}{{{\varepsilon _0}A}}d$ .
Simplifying, we get $C = \dfrac{{{\varepsilon _0}A}}{d}$ .
This suggests that the capacitance increases if the area of the plates increases keeping the distance between the plates as a constant or the capacitance increases if there is a decrease in the separation of the plates.
A change in the size or shape of the capacitor plates is reflected as a change in the area of the plates.
Therefore, the correct options are A, B and D.
Note: Capacitance depends on the area of the plates, the separation between the plates and the nature of the dielectric material placed between the plates. The thickness of the plates do not contribute to the area of the plates and hence has no effect on the capacitance of the plate.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Young’s Double Slit Experiment Derivation Explained

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Advanced Marks vs Rank 2025 - Predict Your IIT Rank Based on Score

