Question 1

A 0.16 pF parallel-plate capacitor is charged to 10 V.Then the battery is disconnected from the capacitor.When 1.00 *107 positive charges of magnitude |e| are now placed on the positive plate of the capacitor, the voltage between the plates changes by:
A)(-5.0 V.)
B)(-1.1 V.)
C)0 V.
D)+1.1 V.
E)+5.0 V.

Accepted Answer

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Question 2

A 200-volt battery is connected to a 0.50-microfarad parallel-plate, air-filled capacitor.Now the battery is disconnected, with care taken not to discharge the plates.Some Pyrex glass is then inserted between the plates, completely filling up the space.What is the final potential difference between the plates? (The dielectric constant for Pyrex is $\kappa$= 5.6.)

Accepted Answer

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Question 3

Into the gap between the plates of a parallel plate capacitor of capacitance a slab of metal is inserted halfway between the plates filling one fourth of the gap between the plates.What is the resulting new capacitance? 
A) 
B) 
C) 
D) 
E)

Accepted Answer

When the slab of metal is inserted halfway between the plates, the distance between the plates is reduced by half. The capacitance is inversely proportional to the distance between the plates, so the capacitance is doubled. However, the slab of metal also acts as a capacitor with the remaining gap between the slab and the plate, adding a capacitance in series with the original capacitor. This effectively halves the capacitance, so the resulting capacitance is the average of the original capacitance and twice the half capacitance. Mathematically, the resulting capacitance can be expressed as:

C = (C/2 + 2C)/(1 + 2) = (5/4)C

where C is the original capacitance.

Therefore, the resulting new capacitance is 5/4 times the original capacitance, which corresponds to choice B.

Question 4

The plates of a parallel plate capacitor of capacitance are horizontal.Into the gap a slab of dielectric material with is placed, filling the bottom half of the gap between the plates.What is the resulting new capacitance?  
A) 
B) 
C) 
D) 
E)

Accepted Answer

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Question 5

A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them.When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates.It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3.After the dielectric is added, the magnitudes of the capacitance and the potential difference between the plates are:
A)C0, V0.  
B)C0, V0. 
C)C0, V0.
D)3C0, V0. 
E)3C0, 3V0.

Accepted Answer

The capacitance of a parallel plate capacitor is given by:
C = εA/d
where ε is the permittivity of the medium between the plates, A is the area of the plates, and d is the distance between them.

When the dielectric material of constant k = 3 is introduced, the capacitance becomes:
C' = kεA/d = 3εA/d = 3C
So the capacitance increases by a factor of 3.

The potential difference between the plates remains the same, as the battery was disconnected and the charge on the plates remained the same. Therefore, the answer is D) 3C0, V0.

Question 6

An initially uncharged parallel plate capacitor of capacitance C is charged to potential V by a battery.The battery is then disconnected.Which statement is correct?
A)There is no charge on either plate of the capacitor.
B)The capacitor can be discharged by grounding any one of its two plates.
C)Charge is distributed evenly over both the inner and outer surfaces of the plates.
D)The magnitude of the electric field outside the space between the plates is approximately zero.
E)The capacitance increases when the distance between the plates increases.

Accepted Answer

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Question 7

Which of the following materials has the highest dielectric constant?
A)air
B)Mylar
C)paper
D)Pyrex glass
E)water

Accepted Answer

Water has the highest dielectric constant among the given materials.

Question 8

An electric dipole having dipole moment of magnitude p is placed in a uniform electric field having magnitude E.What is the magnitude of the greatest change in potential energy that can happen for this dipole in this field?
A)pE
B) 
C)4pE
D) 
E)No answer given is correct.

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Question 9

An electron is released from rest at the negative plate of a parallel plate capacitor.If the distance between the plates is 5 mm and the potential difference across the plates is 5 V, with what velocity does the electron hit the positive plate? (me = 9.1 *10-31 kg, qe = 1.6 * 10-19 C.)

Accepted Answer

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Question 10

Is it feasible to construct an air-filled parallel-plate capacitor that has its two plates separated by 0.10 mm and has a capacitance of 1.0 F? Why or why not?

Accepted Answer

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Question 11

Regarding the Earth and a cloud layer 800 m above the Earth as the 'plates' of a capacitor, calculate the capacitance if the cloud layer has an area of 1.0 km2.If an electric field of 2.0 * 106 N/C makes the air break down and conduct electricity (lightning), what is the maximum charge the cloud can hold?

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Question 12

A 0.120 pF parallel-plate capacitor is charged to a potential difference of 10.0 V and then disconnected from the battery.A cosmic ray burst creates 1.00 * 106 electrons and 1.00 * 106 positive charges between the plates.If the charges do not recombine, but reach the oppositely charged plates, by how much is the potential difference between the capacitor plates reduced?
A)1.33 V
B)7.34 V
C)8.67 V
D)1330 V
E)8670 V

Accepted Answer

The charge added by the cosmic ray burst is 1.6 * 10^-13 C (since 1 electron charge is 1.6 * 10^-19 C). The initial charge on the capacitor is 1.2 * 10^-12 C (C = Q/V, Q = 0.120 pF * 10 V). The new charge is the initial charge minus the charge added by the cosmic rays. The new potential difference is calculated using V = Q/C, resulting in a reduction of 1.33 V.

Question 13

Which of the following statements is incorrect?
A)Capacitance is always positive.
B)The symbol for potential difference between the plates of a capacitor is 
C)Water is a polar molecule.
D)When a dielectric is placed in a capacitor it serves to reduce the electric field.
E)Non-polar molecules cannot be used for dielectric material in a capacitor.

Accepted Answer

The answer of Which of the following statements is incorrect?
A)Capacitance...

Question 14

A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them.When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates.It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3.After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are:
A)Q0, V0.  
B)Q0, V0. 
C)Q0, V0.
D)Q0, 3V0.
E)3Q0, 3V0.

Accepted Answer

When the dielectric is added, the capacitance of the capacitor increases. Capacitance C is given by:
C = ε0 * k * A / d,
where ε0 is the permittivity of free space, k is the dielectric constant, A is the area of the plates, and d is the separation between them.
Since the dielectric constant k increases by a factor of 3, the capacitance C increases by a factor of 3. Therefore, the charge Q on the plates remains the same, but the potential difference V between the plates decreases by a factor of 3. So, the magnitudes of the charge on the plates and the potential difference between them are Q0 and V0/3, respectively.

Question 15

A parallel plate capacitor is connected to a battery and charged to voltage V.Leah says that the charge on the plates will decrease if the distance between the plates is increased while they are still connected to the battery.Gertie says that the charge will remain the same.Which one, if either, is correct, and why?
A)Gertie, because the maximum voltage is determined by the battery.
B)Gertie, because the capacitance of the capacitor does not change.
C)Leah, because the capacitance decreases when the plate separation is increased.
D)Leah, because the capacitance increases when the plate separation is increased.
E)Neither, because the charge increases when the plate separation is increased.

Accepted Answer

C) Leah is correct because the capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$, where $C$ is the capacitance, $\varepsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the distance between the plates. Increasing the distance ($d$) between the plates decreases the capacitance. Since the capacitor is connected to a battery, the voltage ($V$) across the capacitor remains constant. The charge ($Q$) on the capacitor is given by $Q = CV$. Therefore, if the capacitance ($C$) decreases and the voltage ($V$) remains constant, the charge ($Q$) on the plates must decrease.

Question 16

A parallel plate capacitor is charged to voltage V and then disconnected from the battery.Leopold says that the voltage will decrease if the plates are pulled apart.Gerhardt says that the voltage will remain the same.Which one, if either, is correct, and why?
A)Gerhardt, because the maximum voltage is determined by the battery.
B)Gerhardt, because the charge per unit area on the plates does not change.
C)Leopold, because charge is transferred from one plate to the other when the plates are separated.
D)Leopold, because the force each plate exerts on the other decreases when the plates are pulled apart.
E)Neither, because the voltage increases when the plates are pulled apart.

Accepted Answer

When the plates of a disconnected parallel plate capacitor are pulled apart, the capacitance decreases because capacitance is inversely proportional to the distance between the plates (C = εA/d, where A is the area of the plates, d is the distance between them, and ε is the permittivity of the medium between the plates). Since the charge (Q) on the capacitor remains constant (because it's disconnected from the battery), and the voltage (V) across a capacitor is given by V = Q/C, if C decreases, V must increase.

Question 17

Two spheres are made of conducting material.Sphere #2 has twice the radius of Sphere #1.What is the ratio of the capacitance of Sphere #2 to the capacitance of sphere #1?
A)1, since all conducting spheres have the same capacitance.
B)2
C)4
D)8
E)A single sphere has no capacitance since a second concentric spherical shell is necessary to make a spherical capacitor.Thus, none of the answers above is correct.

Accepted Answer

The capacitance of a conducting sphere is given by $C = 4\pi\epsilon_0 r$, where $r$ is the radius of the sphere. If Sphere #2 has twice the radius of Sphere #1, its capacitance is twice as much, since capacitance is directly proportional to the radius.

Question 18

A 0.16 pF parallel-plate capacitor is charged to 10 V.Then the battery is disconnected from the capacitor.When 1.00 * 107 electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by:
A)(-5.0 V.)
B)(-1.1 V.)
C)0 V.
D)+1.1 V.
E)+5.0 V.

Accepted Answer

The answer of A 0.16 pF parallel-plate capacitor is charged...

Quiz 4: Energy and Capacitance

A 0.16 pF parallel-plate capacitor is charged to 10 V.Then the battery is disconnected from the capacitor.When 1.00 *107 positive charges of magnitude |e| are now placed on the positive plate of the capacitor, the voltage between the plates changes by:

Into the gap between the plates of a parallel plate capacitor of capacitance a slab of metal is inserted halfway between the plates filling one fourth of the gap between the plates.What is the resulting new capacitance?

The plates of a parallel plate capacitor of capacitance are horizontal.Into the gap a slab of dielectric material with is placed, filling the bottom half of the gap between the plates.What is the resulting new capacitance?

An initially uncharged parallel plate capacitor of capacitance C is charged to potential V by a battery.The battery is then disconnected.Which statement is correct?

Which of the following materials has the highest dielectric constant?

An electric dipole having dipole moment of magnitude p is placed in a uniform electric field having magnitude E.What is the magnitude of the greatest change in potential energy that can happen for this dipole in this field?

An electron is released from rest at the negative plate of a parallel plate capacitor.If the distance between the plates is 5 mm and the potential difference across the plates is 5 V, with what velocity does the electron hit the positive plate? (me = 9.1 10-31 kg, qe = 1.6 10-19 C.)

Is it feasible to construct an air-filled parallel-plate capacitor that has its two plates separated by 0.10 mm and has a capacitance of 1.0 F? Why or why not?

Which of the following statements is incorrect?

A parallel plate capacitor is charged to voltage V and then disconnected from the battery.Leopold says that the voltage will decrease if the plates are pulled apart.Gerhardt says that the voltage will remain the same.Which one, if either, is correct, and why?

Two spheres are made of conducting material.Sphere #2 has twice the radius of Sphere #1.What is the ratio of the capacitance of Sphere #2 to the capacitance of sphere #1?

A 0.16 pF parallel-plate capacitor is charged to 10 V.Then the battery is disconnected from the capacitor.When 1.00 * 107 electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by: