A parallel plate capacitor is made of two circular plates separated by a distance of \(5~\text{mm}\) and with a dielectric of dielectric constant \(2.2\) between them. When the electric field in the dielectric is \(3\times 10^4~\text{V/m}\), the charge density of the positive plate will be close to:
1. \( 3 \times 10^{-7} ~\text{C} / \text{m}^2 \)
2. \( 3 \times 10^4~\text{C} / \text{m}^2 \)
3. \( 6 \times 10^4 ~\text{C} / \text{m}^2 \)
4. \( 6 \times 10^{-7}~\text{C} / \text{m}^2 \)

Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.

In the given circuit, charge \(Q_2\) on the \(2~\mu\text{F}\) capacitor changes as \(C\) is varied from \(1~\mu\text{F}\) to \(3~\mu\text{F}\)\(Q_2\) as a function of '\(C\)' is given properly by: (figures are drawn schematically and are not to scale)

           

1. 2.
3.   4.  
Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.

The figure shows a system of four capacitors connected across a \(10~\text{V}\) battery. The charge that will flow from the switch \({S}\) when it is closed is: 
 
1. \(5~\mu\text{C}\) from \({b}\) to \({a}\)
2. \({20}~\mu\text{C}\) from \({a}\) to \({b}\)
3. zero
4. \(5~\mu\text{C}\) from \({a}\) to \({b}\)
Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.

advertisementadvertisement

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge \(Q\) (having a charge equal to the sum of the charges on the \(4~\mu \text{F}\) and \(9~\mu\text{F}\) capacitors), at a point distant \(30~\text{m}\) from it, would equal:
                                       
1. \(240~\text{N/C}\)
2. \(360~\text{N/C}\)
3. \(420~\text{N/C}\)
4. \(480~\text{N/C}\)

Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.

A capacitance of \(2~\mu\text{F}\) is required in an electrical circuit across a potential difference of \(1.0~\text{kV}\). A large number of \(1~\mu\text{F}\) capacitors are available which can withstand a potential difference of not more than \(300~\text{V}\). The minimum number of capacitors required to achieve this is:
1. \(2\)
2. \(16\)
3. \(24\)
4. \(32\)

Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints

In the following circuit, the switch \(S\) is closed at \(t = 0.\) The charge on the capacitor \(C_1\) as a function of time will be given by: 
\(\bigg(\text{take}~C_{eq} = \frac{C_1C_2}{C_1C_2}\bigg)\)

1. \(C_1E[1-e^{(-tR/C_1)}] \)
2. \(C_{eq}E~ e^{(-t/RC_{eq})}\)
3. \(C_{eq}E[1-e^{(-t/RC_{eq})}]\)
4. \(C_2E[1-e^{(-t/RC_2)}]\)
Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints

advertisementadvertisement

A parallel plate capacitor has \(1~ \mu\text{F}\) capacitance. One of its two plates is given \(+2~ \mu\text{C}\) charge and the other plate, \(+4~ \mu\text{C}\) charge. The potential difference developed across the capacitor is:
1. \(1~\text{V}\)
2. \(2~\text{V}\) 
3. \(3~\text{V}\)
4. \(5~\text{V}\)

Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.

In the given circuit, the charge on \(4~\mu \text{F}\) capacitor will be:

                     
1. \(5.4~\mu \text{C}\)
2. \(9.6~\mu \text{C}\)
3. \(13.4~\mu \text{C}\)
4. \(24~\mu \text{C}\)

Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.

A parallel plate capacitor is made of two square plates of side \('a',\) separated by a distance \({d(d<<a).} \) The lower triangular portion is filled with a dielectric of dielectric constant \(K,\) as shown in the figure. The capacitance of this capacitor is:
  
1. \(\dfrac{K\varepsilon_0a^2}{2d(K+1)}\)

2. \(\dfrac{K\varepsilon_0a^2 }{ d(K-1)}\mathrm{ln}~K\)

3. \(\dfrac{K\varepsilon_0a^2}{ d}\mathrm{ln}~K\)

4. \(\dfrac{ K\varepsilon_0a^2 }{2d}\)
Subtopic:  Capacitance |
From NCERT
JEE
Please attempt this question first.
Hints

advertisementadvertisement

A \(10~\mu\text{F}\) capacitor is fully charged to a potential difference of \(50\) V. After disconnecting the voltage source, it is connected in parallel with an initially uncharged capacitor. The potential difference across both capacitors becomes \(20\) V. The capacitance of the second capacitor is:

1. \(10~\mu\text{F}\) 2. \(15~\mu\text{F}\)
3. \(20~\mu\text{F}\) 4. \(30~\mu\text{F}\)
Subtopic:  Capacitance |
 75%
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.