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Capacitance

165666 A force of $10 N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

1

5 N

2

10 N

3

20 N

4 Zero

Explanation:

: Given,
We know that, $\begin{array}{r} F = 10 N \\ FE \end{array}$
$q (\frac{Q}{A ε_{0}}) = \frac{qQ}{A ε_{0}}$
$10 = \frac{qQ}{A ε_{0}}$
Now when one plate of capacitor is removed
$E^{'} = \frac{Q}{2 A ε_{o}}$
$F = qE E^{'} = q (\frac{Q}{2 A ε_{0}}) = \frac{1}{2} \frac{qQ}{A ε_{0}}$
$F = \frac{1}{2} \times 10 [(From (i)]$
$F = 5 N$

[JEE Main-27.06.2022

Capacitance

165667 Two metallic plates form a parallel plate capacitor. The distance between the plate is ' $d$ '. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

1

2 : 1

2

1 : 2

3

1 : 4

4

4 : 1

Explanation:

: The capacitance of the air filled parallel plate capacitor is,
$C = \frac{ε_{0} A}{d}$
New capacitance is given by,
$C^{'} = \frac{ε_{0} A}{d - t (1 - \frac{1}{K})}$
After inserting metal sheet of thickness $\frac{d}{2}$ ,
$t = \frac{d}{2}, K = \infty (for metal)$
$or, \frac{1}{K} = 0$
$C^{'} = \frac{ε_{0} A}{d - \frac{d}{2}} = \frac{2 ε_{0} A}{d}$
Hence, from equation (i) \& (ii), we get-
$\frac{C^{'}}{C} = \frac{\frac{2 ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{2}{1}$
$= 2 : 1$

[JEE Main-25.06.2022

Capacitance

165668 A parallel plate air capacitor has a capacitance of $8 pF$ . If the separation between the plates is reduced by half and the space is filled up with a material of dielectric constant 6 , then its capacitance will be.

1

69 pF

2

86 pF

3

68 pF

4

96 pF

Explanation:

: Given capacitance of capacitor,
$C = \frac{ε_{0} A}{d} C = 8 pF$
$8 = \frac{ε_{0} A}{d}$
Now, $C^{'} = \frac{ε_{0} KA}{d^{'}} = \frac{ε_{0} \times 6 \times A}{d / 2} = \frac{12 A ε_{0}}{d}$
$C^{'} = 12 \times 8 \times 10^{- 12} = 96 pF$
$C^{'} = 96 pF$

[Assam CEE-2021]

Capacitance

165669 Three infinite long plane sheets of uniform charge densities $σ_{1} = - 5 σ, σ_{2} = + 2 σ$ and $σ_{1} = 3 σ$ are placed parallel to each other as shown in the figure. The electric field at point $P$ will be

1 Zero

2

- \frac{3 σ}{ε_{0}} \hat{j}

3

- \frac{5 σ}{ε_{0}} \hat{j}

4

- \frac{7 σ}{2 ε_{0}} \hat{j}

Explanation:

: Electric field at point $P =$
$= (\frac{σ_{1}}{2 ε_{0}} + \frac{σ_{2}}{2 ε_{0}} + \frac{σ_{3}}{2 ε_{0}}) \times (- \hat{k})$
$= (\frac{- 5 σ}{2 ε_{0}} + \frac{2 σ}{2 ε_{0}} + \frac{3 σ}{2 ε_{0}}) \times (- \hat{k})$
$= 0$

[Assam CEE-2016]

Capacitance

165666 A force of $10 N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

1

5 N

2

10 N

3

20 N

4 Zero

Explanation:

: Given,
We know that, $\begin{array}{r} F = 10 N \\ FE \end{array}$
$q (\frac{Q}{A ε_{0}}) = \frac{qQ}{A ε_{0}}$
$10 = \frac{qQ}{A ε_{0}}$
Now when one plate of capacitor is removed
$E^{'} = \frac{Q}{2 A ε_{o}}$
$F = qE E^{'} = q (\frac{Q}{2 A ε_{0}}) = \frac{1}{2} \frac{qQ}{A ε_{0}}$
$F = \frac{1}{2} \times 10 [(From (i)]$
$F = 5 N$

[JEE Main-27.06.2022

Capacitance

165667 Two metallic plates form a parallel plate capacitor. The distance between the plate is ' $d$ '. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

1

2 : 1

2

1 : 2

3

1 : 4

4

4 : 1

Explanation:

: The capacitance of the air filled parallel plate capacitor is,
$C = \frac{ε_{0} A}{d}$
New capacitance is given by,
$C^{'} = \frac{ε_{0} A}{d - t (1 - \frac{1}{K})}$
After inserting metal sheet of thickness $\frac{d}{2}$ ,
$t = \frac{d}{2}, K = \infty (for metal)$
$or, \frac{1}{K} = 0$
$C^{'} = \frac{ε_{0} A}{d - \frac{d}{2}} = \frac{2 ε_{0} A}{d}$
Hence, from equation (i) \& (ii), we get-
$\frac{C^{'}}{C} = \frac{\frac{2 ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{2}{1}$
$= 2 : 1$

[JEE Main-25.06.2022

Capacitance

165668 A parallel plate air capacitor has a capacitance of $8 pF$ . If the separation between the plates is reduced by half and the space is filled up with a material of dielectric constant 6 , then its capacitance will be.

1

69 pF

2

86 pF

3

68 pF

4

96 pF

Explanation:

: Given capacitance of capacitor,
$C = \frac{ε_{0} A}{d} C = 8 pF$
$8 = \frac{ε_{0} A}{d}$
Now, $C^{'} = \frac{ε_{0} KA}{d^{'}} = \frac{ε_{0} \times 6 \times A}{d / 2} = \frac{12 A ε_{0}}{d}$
$C^{'} = 12 \times 8 \times 10^{- 12} = 96 pF$
$C^{'} = 96 pF$

[Assam CEE-2021]

Capacitance

165669 Three infinite long plane sheets of uniform charge densities $σ_{1} = - 5 σ, σ_{2} = + 2 σ$ and $σ_{1} = 3 σ$ are placed parallel to each other as shown in the figure. The electric field at point $P$ will be

1 Zero

2

- \frac{3 σ}{ε_{0}} \hat{j}

3

- \frac{5 σ}{ε_{0}} \hat{j}

4

- \frac{7 σ}{2 ε_{0}} \hat{j}

Explanation:

: Electric field at point $P =$
$= (\frac{σ_{1}}{2 ε_{0}} + \frac{σ_{2}}{2 ε_{0}} + \frac{σ_{3}}{2 ε_{0}}) \times (- \hat{k})$
$= (\frac{- 5 σ}{2 ε_{0}} + \frac{2 σ}{2 ε_{0}} + \frac{3 σ}{2 ε_{0}}) \times (- \hat{k})$
$= 0$

[Assam CEE-2016]

Capacitance

165666 A force of $10 N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

1

5 N

2

10 N

3

20 N

4 Zero

Explanation:

: Given,
We know that, $\begin{array}{r} F = 10 N \\ FE \end{array}$
$q (\frac{Q}{A ε_{0}}) = \frac{qQ}{A ε_{0}}$
$10 = \frac{qQ}{A ε_{0}}$
Now when one plate of capacitor is removed
$E^{'} = \frac{Q}{2 A ε_{o}}$
$F = qE E^{'} = q (\frac{Q}{2 A ε_{0}}) = \frac{1}{2} \frac{qQ}{A ε_{0}}$
$F = \frac{1}{2} \times 10 [(From (i)]$
$F = 5 N$

[JEE Main-27.06.2022

Capacitance

165667 Two metallic plates form a parallel plate capacitor. The distance between the plate is ' $d$ '. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

1

2 : 1

2

1 : 2

3

1 : 4

4

4 : 1

Explanation:

: The capacitance of the air filled parallel plate capacitor is,
$C = \frac{ε_{0} A}{d}$
New capacitance is given by,
$C^{'} = \frac{ε_{0} A}{d - t (1 - \frac{1}{K})}$
After inserting metal sheet of thickness $\frac{d}{2}$ ,
$t = \frac{d}{2}, K = \infty (for metal)$
$or, \frac{1}{K} = 0$
$C^{'} = \frac{ε_{0} A}{d - \frac{d}{2}} = \frac{2 ε_{0} A}{d}$
Hence, from equation (i) \& (ii), we get-
$\frac{C^{'}}{C} = \frac{\frac{2 ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{2}{1}$
$= 2 : 1$

[JEE Main-25.06.2022

Capacitance

165668 A parallel plate air capacitor has a capacitance of $8 pF$ . If the separation between the plates is reduced by half and the space is filled up with a material of dielectric constant 6 , then its capacitance will be.

1

69 pF

2

86 pF

3

68 pF

4

96 pF

Explanation:

: Given capacitance of capacitor,
$C = \frac{ε_{0} A}{d} C = 8 pF$
$8 = \frac{ε_{0} A}{d}$
Now, $C^{'} = \frac{ε_{0} KA}{d^{'}} = \frac{ε_{0} \times 6 \times A}{d / 2} = \frac{12 A ε_{0}}{d}$
$C^{'} = 12 \times 8 \times 10^{- 12} = 96 pF$
$C^{'} = 96 pF$

[Assam CEE-2021]

Capacitance

165669 Three infinite long plane sheets of uniform charge densities $σ_{1} = - 5 σ, σ_{2} = + 2 σ$ and $σ_{1} = 3 σ$ are placed parallel to each other as shown in the figure. The electric field at point $P$ will be

1 Zero

2

- \frac{3 σ}{ε_{0}} \hat{j}

3

- \frac{5 σ}{ε_{0}} \hat{j}

4

- \frac{7 σ}{2 ε_{0}} \hat{j}

Explanation:

: Electric field at point $P =$
$= (\frac{σ_{1}}{2 ε_{0}} + \frac{σ_{2}}{2 ε_{0}} + \frac{σ_{3}}{2 ε_{0}}) \times (- \hat{k})$
$= (\frac{- 5 σ}{2 ε_{0}} + \frac{2 σ}{2 ε_{0}} + \frac{3 σ}{2 ε_{0}}) \times (- \hat{k})$
$= 0$

[Assam CEE-2016]

Capacitance

165666 A force of $10 N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

1

5 N

2

10 N

3

20 N

4 Zero

Explanation:

: Given,
We know that, $\begin{array}{r} F = 10 N \\ FE \end{array}$
$q (\frac{Q}{A ε_{0}}) = \frac{qQ}{A ε_{0}}$
$10 = \frac{qQ}{A ε_{0}}$
Now when one plate of capacitor is removed
$E^{'} = \frac{Q}{2 A ε_{o}}$
$F = qE E^{'} = q (\frac{Q}{2 A ε_{0}}) = \frac{1}{2} \frac{qQ}{A ε_{0}}$
$F = \frac{1}{2} \times 10 [(From (i)]$
$F = 5 N$

[JEE Main-27.06.2022

Capacitance

165667 Two metallic plates form a parallel plate capacitor. The distance between the plate is ' $d$ '. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

1

2 : 1

2

1 : 2

3

1 : 4

4

4 : 1

Explanation:

: The capacitance of the air filled parallel plate capacitor is,
$C = \frac{ε_{0} A}{d}$
New capacitance is given by,
$C^{'} = \frac{ε_{0} A}{d - t (1 - \frac{1}{K})}$
After inserting metal sheet of thickness $\frac{d}{2}$ ,
$t = \frac{d}{2}, K = \infty (for metal)$
$or, \frac{1}{K} = 0$
$C^{'} = \frac{ε_{0} A}{d - \frac{d}{2}} = \frac{2 ε_{0} A}{d}$
Hence, from equation (i) \& (ii), we get-
$\frac{C^{'}}{C} = \frac{\frac{2 ε_{0} A}{d}}{\frac{ε_{0} A}{d}} = \frac{2}{1}$
$= 2 : 1$

[JEE Main-25.06.2022

Capacitance

165668 A parallel plate air capacitor has a capacitance of $8 pF$ . If the separation between the plates is reduced by half and the space is filled up with a material of dielectric constant 6 , then its capacitance will be.

1

69 pF

2

86 pF

3

68 pF

4

96 pF

Explanation:

: Given capacitance of capacitor,
$C = \frac{ε_{0} A}{d} C = 8 pF$
$8 = \frac{ε_{0} A}{d}$
Now, $C^{'} = \frac{ε_{0} KA}{d^{'}} = \frac{ε_{0} \times 6 \times A}{d / 2} = \frac{12 A ε_{0}}{d}$
$C^{'} = 12 \times 8 \times 10^{- 12} = 96 pF$
$C^{'} = 96 pF$

[Assam CEE-2021]

Capacitance

165669 Three infinite long plane sheets of uniform charge densities $σ_{1} = - 5 σ, σ_{2} = + 2 σ$ and $σ_{1} = 3 σ$ are placed parallel to each other as shown in the figure. The electric field at point $P$ will be

1 Zero

2

- \frac{3 σ}{ε_{0}} \hat{j}

3

- \frac{5 σ}{ε_{0}} \hat{j}

4

- \frac{7 σ}{2 ε_{0}} \hat{j}

Explanation:

: Electric field at point $P =$
$= (\frac{σ_{1}}{2 ε_{0}} + \frac{σ_{2}}{2 ε_{0}} + \frac{σ_{3}}{2 ε_{0}}) \times (- \hat{k})$
$= (\frac{- 5 σ}{2 ε_{0}} + \frac{2 σ}{2 ε_{0}} + \frac{3 σ}{2 ε_{0}}) \times (- \hat{k})$
$= 0$

[Assam CEE-2016]