QBManager

Capacitance

165694 The four capacitors, each of $25 μ F$ are connected as shown in figure. The DC voltmeter reads $200 V$ . The charge on each plate of capacitor is

1

\pm 2 \times 10^{- 3} C

2

\pm 5 \times 10^{- 3} C

3

\pm 2 \times 10^{- 2} C

4

\pm 5 \times 10^{- 2} C

Explanation:

:

Two capacitors on the left side are in parallel and potential across each equal to $V$
Similarly, the capacitors on right side are in parallel and potential across each equal to $V$ .
Therefore, charge on each plate of capacitor is,
$Q = \pm 25 \times 10^{- 6} \times 200$
$Q = \pm 5 \times 10^{- 3} C$

[AIPMT-1994]

Capacitance

165696 A parallel plate condenser has a uniform electric field $E (V / m)$ in the space between the plates. If the distance between the plates is $d (m)$ and area of each plate is $A (m^{2})$ . The energy (joule) stored in the condenser is

1

\frac{1}{2} ε_{0} E^{2}

2

ε_{0}

EAd (c)

3

\frac{1}{2} ε_{0} E^{2} Ad

4

E^{2} Ad / ε_{0}

Explanation:

: The capacitance of parallel plate condense is
$C = \frac{A ε_{0}}{d}$
And potential difference between the plates is $V = Ed$
$∴$ Energy stored
$U = \frac{1}{2} {CV}^{2} = \frac{1}{2} (\frac{A ε_{0}}{d}) (Ed)^{2}$
$= \frac{1}{2} A ε_{0} E^{2} d$

[NEET-2021

Capacitance

165697 Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf $V$ volt. Now, disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?

1

75 %

2

0 %

3

50 %

4

25 %

Explanation:

: When connected within a \& b across $C_{1}$
$E_{1} = \frac{1}{2} {CV}^{2}$
When connected within $b$ and $c$ across $C_{1}$ and $C_{2}$
$Δ E = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} V^{2}$
$= \frac{1}{2} \frac{C^{2}}{2 C} V^{2}$
$= \frac{1}{4} {CV}^{2}$
$%$ loss in energy $= \frac{Δ E}{E_{1}} \times 100$
$= \frac{\frac{1}{4} {CV}^{2}}{\frac{1}{2} {CV}^{2}} \times 100$
$= 50 %$

[NEET Odisha-2019]

Capacitance

165698 A capacitor of capacitance $6 μ F$ is charged upto $100 V$ . The energy stored in the capacitor is

1

0.6 J

2

0.06 J

3

0.03 J

4

0.3 J

Explanation:

: Given, $C = 6 \times 10^{- 6} F, V = 100 V$
The energy stored in the capacitor is given by,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 6 \times 10^{- 6} \times (100)^{2}$
$= 3 \times 10^{- 2} J$
So, $U = 0.03 J$

[UP CPMT-2004]

Capacitance

165694 The four capacitors, each of $25 μ F$ are connected as shown in figure. The DC voltmeter reads $200 V$ . The charge on each plate of capacitor is

1

\pm 2 \times 10^{- 3} C

2

\pm 5 \times 10^{- 3} C

3

\pm 2 \times 10^{- 2} C

4

\pm 5 \times 10^{- 2} C

Explanation:

:

Two capacitors on the left side are in parallel and potential across each equal to $V$
Similarly, the capacitors on right side are in parallel and potential across each equal to $V$ .
Therefore, charge on each plate of capacitor is,
$Q = \pm 25 \times 10^{- 6} \times 200$
$Q = \pm 5 \times 10^{- 3} C$

[AIPMT-1994]

Capacitance

165696 A parallel plate condenser has a uniform electric field $E (V / m)$ in the space between the plates. If the distance between the plates is $d (m)$ and area of each plate is $A (m^{2})$ . The energy (joule) stored in the condenser is

1

\frac{1}{2} ε_{0} E^{2}

2

ε_{0}

EAd (c)

3

\frac{1}{2} ε_{0} E^{2} Ad

4

E^{2} Ad / ε_{0}

Explanation:

: The capacitance of parallel plate condense is
$C = \frac{A ε_{0}}{d}$
And potential difference between the plates is $V = Ed$
$∴$ Energy stored
$U = \frac{1}{2} {CV}^{2} = \frac{1}{2} (\frac{A ε_{0}}{d}) (Ed)^{2}$
$= \frac{1}{2} A ε_{0} E^{2} d$

[NEET-2021

Capacitance

165697 Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf $V$ volt. Now, disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?

1

75 %

2

0 %

3

50 %

4

25 %

Explanation:

: When connected within a \& b across $C_{1}$
$E_{1} = \frac{1}{2} {CV}^{2}$
When connected within $b$ and $c$ across $C_{1}$ and $C_{2}$
$Δ E = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} V^{2}$
$= \frac{1}{2} \frac{C^{2}}{2 C} V^{2}$
$= \frac{1}{4} {CV}^{2}$
$%$ loss in energy $= \frac{Δ E}{E_{1}} \times 100$
$= \frac{\frac{1}{4} {CV}^{2}}{\frac{1}{2} {CV}^{2}} \times 100$
$= 50 %$

[NEET Odisha-2019]

Capacitance

165698 A capacitor of capacitance $6 μ F$ is charged upto $100 V$ . The energy stored in the capacitor is

1

0.6 J

2

0.06 J

3

0.03 J

4

0.3 J

Explanation:

: Given, $C = 6 \times 10^{- 6} F, V = 100 V$
The energy stored in the capacitor is given by,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 6 \times 10^{- 6} \times (100)^{2}$
$= 3 \times 10^{- 2} J$
So, $U = 0.03 J$

[UP CPMT-2004]

Capacitance

165694 The four capacitors, each of $25 μ F$ are connected as shown in figure. The DC voltmeter reads $200 V$ . The charge on each plate of capacitor is

1

\pm 2 \times 10^{- 3} C

2

\pm 5 \times 10^{- 3} C

3

\pm 2 \times 10^{- 2} C

4

\pm 5 \times 10^{- 2} C

Explanation:

:

Two capacitors on the left side are in parallel and potential across each equal to $V$
Similarly, the capacitors on right side are in parallel and potential across each equal to $V$ .
Therefore, charge on each plate of capacitor is,
$Q = \pm 25 \times 10^{- 6} \times 200$
$Q = \pm 5 \times 10^{- 3} C$

[AIPMT-1994]

Capacitance

165696 A parallel plate condenser has a uniform electric field $E (V / m)$ in the space between the plates. If the distance between the plates is $d (m)$ and area of each plate is $A (m^{2})$ . The energy (joule) stored in the condenser is

1

\frac{1}{2} ε_{0} E^{2}

2

ε_{0}

EAd (c)

3

\frac{1}{2} ε_{0} E^{2} Ad

4

E^{2} Ad / ε_{0}

Explanation:

: The capacitance of parallel plate condense is
$C = \frac{A ε_{0}}{d}$
And potential difference between the plates is $V = Ed$
$∴$ Energy stored
$U = \frac{1}{2} {CV}^{2} = \frac{1}{2} (\frac{A ε_{0}}{d}) (Ed)^{2}$
$= \frac{1}{2} A ε_{0} E^{2} d$

[NEET-2021

Capacitance

165697 Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf $V$ volt. Now, disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?

1

75 %

2

0 %

3

50 %

4

25 %

Explanation:

: When connected within a \& b across $C_{1}$
$E_{1} = \frac{1}{2} {CV}^{2}$
When connected within $b$ and $c$ across $C_{1}$ and $C_{2}$
$Δ E = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} V^{2}$
$= \frac{1}{2} \frac{C^{2}}{2 C} V^{2}$
$= \frac{1}{4} {CV}^{2}$
$%$ loss in energy $= \frac{Δ E}{E_{1}} \times 100$
$= \frac{\frac{1}{4} {CV}^{2}}{\frac{1}{2} {CV}^{2}} \times 100$
$= 50 %$

[NEET Odisha-2019]

Capacitance

165698 A capacitor of capacitance $6 μ F$ is charged upto $100 V$ . The energy stored in the capacitor is

1

0.6 J

2

0.06 J

3

0.03 J

4

0.3 J

Explanation:

: Given, $C = 6 \times 10^{- 6} F, V = 100 V$
The energy stored in the capacitor is given by,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 6 \times 10^{- 6} \times (100)^{2}$
$= 3 \times 10^{- 2} J$
So, $U = 0.03 J$

[UP CPMT-2004]

Capacitance

165694 The four capacitors, each of $25 μ F$ are connected as shown in figure. The DC voltmeter reads $200 V$ . The charge on each plate of capacitor is

1

\pm 2 \times 10^{- 3} C

2

\pm 5 \times 10^{- 3} C

3

\pm 2 \times 10^{- 2} C

4

\pm 5 \times 10^{- 2} C

Explanation:

:

Two capacitors on the left side are in parallel and potential across each equal to $V$
Similarly, the capacitors on right side are in parallel and potential across each equal to $V$ .
Therefore, charge on each plate of capacitor is,
$Q = \pm 25 \times 10^{- 6} \times 200$
$Q = \pm 5 \times 10^{- 3} C$

[AIPMT-1994]

Capacitance

165696 A parallel plate condenser has a uniform electric field $E (V / m)$ in the space between the plates. If the distance between the plates is $d (m)$ and area of each plate is $A (m^{2})$ . The energy (joule) stored in the condenser is

1

\frac{1}{2} ε_{0} E^{2}

2

ε_{0}

EAd (c)

3

\frac{1}{2} ε_{0} E^{2} Ad

4

E^{2} Ad / ε_{0}

Explanation:

: The capacitance of parallel plate condense is
$C = \frac{A ε_{0}}{d}$
And potential difference between the plates is $V = Ed$
$∴$ Energy stored
$U = \frac{1}{2} {CV}^{2} = \frac{1}{2} (\frac{A ε_{0}}{d}) (Ed)^{2}$
$= \frac{1}{2} A ε_{0} E^{2} d$

[NEET-2021

Capacitance

165697 Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf $V$ volt. Now, disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?

1

75 %

2

0 %

3

50 %

4

25 %

Explanation:

: When connected within a \& b across $C_{1}$
$E_{1} = \frac{1}{2} {CV}^{2}$
When connected within $b$ and $c$ across $C_{1}$ and $C_{2}$
$Δ E = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} V^{2}$
$= \frac{1}{2} \frac{C^{2}}{2 C} V^{2}$
$= \frac{1}{4} {CV}^{2}$
$%$ loss in energy $= \frac{Δ E}{E_{1}} \times 100$
$= \frac{\frac{1}{4} {CV}^{2}}{\frac{1}{2} {CV}^{2}} \times 100$
$= 50 %$

[NEET Odisha-2019]

Capacitance

165698 A capacitor of capacitance $6 μ F$ is charged upto $100 V$ . The energy stored in the capacitor is

1

0.6 J

2

0.06 J

3

0.03 J

4

0.3 J

Explanation:

: Given, $C = 6 \times 10^{- 6} F, V = 100 V$
The energy stored in the capacitor is given by,
$U = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 6 \times 10^{- 6} \times (100)^{2}$
$= 3 \times 10^{- 2} J$
So, $U = 0.03 J$

[UP CPMT-2004]

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: