
A parallel plate capacitor is charged and then isolated. What is the effect of increasing the plate separation on charge, potential, capacitance, respectively?
(A) Constant, decreases, decreases
(B) Increases, decreases, decreases
(C) Constant, decreases, increases
(D) Constant. Increases, decreases
Answer
544.5k+ views
Hint:Capacitance in general is defined as the ratio of the charge on the plates of the conductor to the potential difference. We can find the effect of increase in the plate separation on charge(Q), potential(V) and capacitance(C). The capacitance of a capacitor depends upon a number of factors but the most important is what type of capacitor it is whether a parallel plate capacitor or cylindrical capacitor.
Formula Used:
Capacitance for a parallel plate capacitor is given as $C=\dfrac{{{\varepsilon }_{0}}A}{d}$,
where A is area of cross section, d is the distance between two plates and ${{\varepsilon }_{0}}$is the relative permittivity.
Complete answer:
Capacitance(C) is defined as the ratio of the charge stored in the conductor to the potential difference(V), $C=Q/V$. Charge on the plates is, $Q=CV$. And the potential between the plates is $V=Q/C$. For a parallel plate capacitor the capacitance is equal to $C=\dfrac{{{\varepsilon }_{0}}A}{d}$, where A is area of cross section, d is the distance between two plates and ${{\varepsilon }_{0}}$ is the relative permittivity.
Therefore with an increase in the separation between the plates (d), capacitance increases as it is inversely related to distance between plates of capacitor and potential decreases as it is inversely proportional to capacitance. Due to the increase in capacitance and decreases in potential by the same factor, the charge remains constant.
Note:One should know the formula for capacitance for a parallel plate capacitor to solve such types of problems. A cylindrical capacitor consists of a hollow or a solid cylindrical conductor surrounded by another concentric hollow spherical cylinder. When the capacitor is connected to a battery then the charging process gets started. The charge moves from one plate of the capacitor to the other and so an electric field is produced in the space between the two plates.
Formula Used:
Capacitance for a parallel plate capacitor is given as $C=\dfrac{{{\varepsilon }_{0}}A}{d}$,
where A is area of cross section, d is the distance between two plates and ${{\varepsilon }_{0}}$is the relative permittivity.
Complete answer:
Capacitance(C) is defined as the ratio of the charge stored in the conductor to the potential difference(V), $C=Q/V$. Charge on the plates is, $Q=CV$. And the potential between the plates is $V=Q/C$. For a parallel plate capacitor the capacitance is equal to $C=\dfrac{{{\varepsilon }_{0}}A}{d}$, where A is area of cross section, d is the distance between two plates and ${{\varepsilon }_{0}}$ is the relative permittivity.
Therefore with an increase in the separation between the plates (d), capacitance increases as it is inversely related to distance between plates of capacitor and potential decreases as it is inversely proportional to capacitance. Due to the increase in capacitance and decreases in potential by the same factor, the charge remains constant.
Note:One should know the formula for capacitance for a parallel plate capacitor to solve such types of problems. A cylindrical capacitor consists of a hollow or a solid cylindrical conductor surrounded by another concentric hollow spherical cylinder. When the capacitor is connected to a battery then the charging process gets started. The charge moves from one plate of the capacitor to the other and so an electric field is produced in the space between the two plates.
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