
The nature of the capacitance of an electrostatic capacitor depends on:
A) Shape
B) Size
C) Thickness of plate
D) Area
Answer
219.3k+ 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.
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