165597
The capacity of a parallel plate capacitor increases with the
1 Decreases of its area
2 Increase of its distance
3 Increases of its area
4 None of these
Explanation:
: We know that, In a parallel plate capacitor, the capacity of capacitor, $\mathrm{C}=\mathrm{K} \frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}$ Where, $\mathrm{A}=$ Area of the plate $\mathrm{d}=$ Distance between two plate Then, $\quad \mathrm{C} \propto \mathrm{A} \quad \& \quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ So, the capacity of a parallel plate capacitor increases with the increase its area and the decrease its distance.
[Karnataka CET-2015
Capacitance
165598
$Q$ charge is given to two capacitors $C_{1}$ and $C_{2}$ which are in parallel. The charge distribution among them is
1 $\mathrm{C}_{1}: \mathrm{C}_{2}$
2 $\mathrm{C}_{2}: \mathrm{C}_{1}$
3 $\mathrm{C}_{1} \mathrm{C}_{2}: 1$
4 $1: \mathrm{C}_{1} \mathrm{C}_{2}$
Explanation:
: When $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel they have the same potential difference. Then, $\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}$ We know, $\mathrm{Q}=\mathrm{C} . \mathrm{V}$ So, $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1} \mathrm{~V}}{\mathrm{C}_{2} \mathrm{~V}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$ $\mathrm{Q}_{1}: \mathrm{Q}_{2}=\mathrm{C}_{1}: \mathrm{C}_{2}$
[CG PET- 2012]
Capacitance
165599
The capacitance of a parallel- plate condenser does not depend upon
1 Area of the plates
2 Medium between the plates
3 Distance between the plates
4 Metal of the plates
Explanation:
: From question, For parallel plate condenser, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where, $A=$ Area of the parallel plate $\mathrm{d}=$ Distance between parallel plate \(\varepsilon_{\mathrm{o}}=\) Permittivity of free space \(\left(8.85 \times 10^{-12}\right.\mathrm{F} / \mathrm{m})\) $\mathrm{K}=\text { Dielectric constant }$ Hence, the capacitance of a parallel plate condenser does not depend upon metal of plates.
[CG PET- 2010]
Capacitance
165600
A $500 \mu \mathrm{F}$ capacitor is charged at a steady rate of $100 \mu \mathrm{C} / \mathrm{s}$. The potential difference across the capacitor will be $10 \mathrm{~V}$ after an interval of
165597
The capacity of a parallel plate capacitor increases with the
1 Decreases of its area
2 Increase of its distance
3 Increases of its area
4 None of these
Explanation:
: We know that, In a parallel plate capacitor, the capacity of capacitor, $\mathrm{C}=\mathrm{K} \frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}$ Where, $\mathrm{A}=$ Area of the plate $\mathrm{d}=$ Distance between two plate Then, $\quad \mathrm{C} \propto \mathrm{A} \quad \& \quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ So, the capacity of a parallel plate capacitor increases with the increase its area and the decrease its distance.
[Karnataka CET-2015
Capacitance
165598
$Q$ charge is given to two capacitors $C_{1}$ and $C_{2}$ which are in parallel. The charge distribution among them is
1 $\mathrm{C}_{1}: \mathrm{C}_{2}$
2 $\mathrm{C}_{2}: \mathrm{C}_{1}$
3 $\mathrm{C}_{1} \mathrm{C}_{2}: 1$
4 $1: \mathrm{C}_{1} \mathrm{C}_{2}$
Explanation:
: When $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel they have the same potential difference. Then, $\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}$ We know, $\mathrm{Q}=\mathrm{C} . \mathrm{V}$ So, $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1} \mathrm{~V}}{\mathrm{C}_{2} \mathrm{~V}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$ $\mathrm{Q}_{1}: \mathrm{Q}_{2}=\mathrm{C}_{1}: \mathrm{C}_{2}$
[CG PET- 2012]
Capacitance
165599
The capacitance of a parallel- plate condenser does not depend upon
1 Area of the plates
2 Medium between the plates
3 Distance between the plates
4 Metal of the plates
Explanation:
: From question, For parallel plate condenser, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where, $A=$ Area of the parallel plate $\mathrm{d}=$ Distance between parallel plate \(\varepsilon_{\mathrm{o}}=\) Permittivity of free space \(\left(8.85 \times 10^{-12}\right.\mathrm{F} / \mathrm{m})\) $\mathrm{K}=\text { Dielectric constant }$ Hence, the capacitance of a parallel plate condenser does not depend upon metal of plates.
[CG PET- 2010]
Capacitance
165600
A $500 \mu \mathrm{F}$ capacitor is charged at a steady rate of $100 \mu \mathrm{C} / \mathrm{s}$. The potential difference across the capacitor will be $10 \mathrm{~V}$ after an interval of
165597
The capacity of a parallel plate capacitor increases with the
1 Decreases of its area
2 Increase of its distance
3 Increases of its area
4 None of these
Explanation:
: We know that, In a parallel plate capacitor, the capacity of capacitor, $\mathrm{C}=\mathrm{K} \frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}$ Where, $\mathrm{A}=$ Area of the plate $\mathrm{d}=$ Distance between two plate Then, $\quad \mathrm{C} \propto \mathrm{A} \quad \& \quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ So, the capacity of a parallel plate capacitor increases with the increase its area and the decrease its distance.
[Karnataka CET-2015
Capacitance
165598
$Q$ charge is given to two capacitors $C_{1}$ and $C_{2}$ which are in parallel. The charge distribution among them is
1 $\mathrm{C}_{1}: \mathrm{C}_{2}$
2 $\mathrm{C}_{2}: \mathrm{C}_{1}$
3 $\mathrm{C}_{1} \mathrm{C}_{2}: 1$
4 $1: \mathrm{C}_{1} \mathrm{C}_{2}$
Explanation:
: When $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel they have the same potential difference. Then, $\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}$ We know, $\mathrm{Q}=\mathrm{C} . \mathrm{V}$ So, $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1} \mathrm{~V}}{\mathrm{C}_{2} \mathrm{~V}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$ $\mathrm{Q}_{1}: \mathrm{Q}_{2}=\mathrm{C}_{1}: \mathrm{C}_{2}$
[CG PET- 2012]
Capacitance
165599
The capacitance of a parallel- plate condenser does not depend upon
1 Area of the plates
2 Medium between the plates
3 Distance between the plates
4 Metal of the plates
Explanation:
: From question, For parallel plate condenser, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where, $A=$ Area of the parallel plate $\mathrm{d}=$ Distance between parallel plate \(\varepsilon_{\mathrm{o}}=\) Permittivity of free space \(\left(8.85 \times 10^{-12}\right.\mathrm{F} / \mathrm{m})\) $\mathrm{K}=\text { Dielectric constant }$ Hence, the capacitance of a parallel plate condenser does not depend upon metal of plates.
[CG PET- 2010]
Capacitance
165600
A $500 \mu \mathrm{F}$ capacitor is charged at a steady rate of $100 \mu \mathrm{C} / \mathrm{s}$. The potential difference across the capacitor will be $10 \mathrm{~V}$ after an interval of
165597
The capacity of a parallel plate capacitor increases with the
1 Decreases of its area
2 Increase of its distance
3 Increases of its area
4 None of these
Explanation:
: We know that, In a parallel plate capacitor, the capacity of capacitor, $\mathrm{C}=\mathrm{K} \frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}$ Where, $\mathrm{A}=$ Area of the plate $\mathrm{d}=$ Distance between two plate Then, $\quad \mathrm{C} \propto \mathrm{A} \quad \& \quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ So, the capacity of a parallel plate capacitor increases with the increase its area and the decrease its distance.
[Karnataka CET-2015
Capacitance
165598
$Q$ charge is given to two capacitors $C_{1}$ and $C_{2}$ which are in parallel. The charge distribution among them is
1 $\mathrm{C}_{1}: \mathrm{C}_{2}$
2 $\mathrm{C}_{2}: \mathrm{C}_{1}$
3 $\mathrm{C}_{1} \mathrm{C}_{2}: 1$
4 $1: \mathrm{C}_{1} \mathrm{C}_{2}$
Explanation:
: When $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel they have the same potential difference. Then, $\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}$ We know, $\mathrm{Q}=\mathrm{C} . \mathrm{V}$ So, $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1} \mathrm{~V}}{\mathrm{C}_{2} \mathrm{~V}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$ $\mathrm{Q}_{1}: \mathrm{Q}_{2}=\mathrm{C}_{1}: \mathrm{C}_{2}$
[CG PET- 2012]
Capacitance
165599
The capacitance of a parallel- plate condenser does not depend upon
1 Area of the plates
2 Medium between the plates
3 Distance between the plates
4 Metal of the plates
Explanation:
: From question, For parallel plate condenser, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where, $A=$ Area of the parallel plate $\mathrm{d}=$ Distance between parallel plate \(\varepsilon_{\mathrm{o}}=\) Permittivity of free space \(\left(8.85 \times 10^{-12}\right.\mathrm{F} / \mathrm{m})\) $\mathrm{K}=\text { Dielectric constant }$ Hence, the capacitance of a parallel plate condenser does not depend upon metal of plates.
[CG PET- 2010]
Capacitance
165600
A $500 \mu \mathrm{F}$ capacitor is charged at a steady rate of $100 \mu \mathrm{C} / \mathrm{s}$. The potential difference across the capacitor will be $10 \mathrm{~V}$ after an interval of
