Question 1

Each plate of an ideal air-filled parallel-plate capacitor has an area of 0.0020 ,and the separation of the plates is An electric field of Is present between the plates.What is the surface charge density on the plates? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Accepted Answer

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Question 2

An ideal air-filled parallel-plate capacitor with horizontal plates has a plate separation of 5.0 cm.If the potential difference between the plates is 2000 V,with the top plate at the higher potential,what are the magnitude and direction of the electric field between the plates?

Accepted Answer

The electric field (E) between the plates of a parallel-plate capacitor is given by E = V/d, where V is the potential difference and d is the separation between the plates. Here, E = 2000 V / 0.05 m = 40000 N/C. The direction is downward because the electric field points from the positive (higher potential) to the negative (lower potential) plate.

Question 3

Two isolated copper plates,each of area 0.40 m²,carry opposite charges of magnitude 7.08 × 10^-10 C.They are placed opposite each other in parallel alignment,with a spacing of 4.0 cm between them.What is the potential difference between the plates? (ε₀ = 8.85 × 10^-12 C²/N ∙ m²)

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Question 4

A proton moves 0.10 m along the direction of an electric field of magnitude 3.0 V/m.What is the change in kinetic energy of the proton? (e = 1.60 × 10^-19C)

Accepted Answer

The work done on a charged particle moving in an electric field is given by W=qEd, where q is the charge of the particle, E is the electric field strength, and d is the distance moved by the particle in the direction of the electric field. 
In this case, the proton has a charge of +1.60 × 10^-19 C and moves a distance of 0.10 m in a field of 3.0 V/m, so the work done on the proton is W = (1.60 × 10^-19 C)(3.0 V/m)(0.10 m) = 4.8 × 10^-20 J.
By the work-energy theorem, the change in kinetic energy of the proton is equal to the work done on it, so the change in kinetic energy is 4.8 × 10^-20 J. Therefore, the best choice is A.

Question 5

An ideal air-filled parallel plate capacitor with plate a separation of 4.0 cm has a plate area of 0.040 m².What is the capacitance of this capacitor with air between these plates? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Accepted Answer

The capacitance of an air-filled parallel plate capacitor is given by the equation C = εA/d, where ε is the permittivity of free space, A is the area of the plate, and d is the distance between the plates. Substituting the given values, we get C = (8.85 × 10^-12 N^-1 m^-2) * (0.040 m^2) / (4.0 cm = 0.04 m) = 8.9 pF. Therefore, the best choice is B.

Question 6

The potential difference between the plates of an ideal air-filled parallel-plate capacitor with a plate separation of 6.0 cm is 60 V.What is the strength of the electric field between the plates of this capacitor?

Accepted Answer

The electric field strength between the plates of a parallel-plate capacitor is given by E = V/d, where V is the potential difference between the plates and d is the distance between the plates. Plugging in the given values, we get:
E = 60 V / 0.06 m = 1000 V/m
Therefore, the strength of the electric field between the plates is 1000 V/m, which is option C.

Question 7

Consider a uniform horizontal electric field of 50 N/C directed toward the east.If the electric potential at a given point in the field is 80 V,what is the potential at a point 1.0 m directly east of the point?

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Question 8

When the magnitude of the charge on each plate of an air-filled capacitor is 4 μC,the potential difference between the plates is 80 V.What is the capacitance of this capacitor?

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Question 9

Two very large parallel metal plates,separated by 0.20 m,are connected across a 12-V source of potential.An electron is released from rest at a location 0.10 m from the negative plate.When the electron arrives at a distance 0.050 m from the positive plate,how much kinetic energy has the electron gained?

Accepted Answer

The electron gains potential energy as it moves from the negative plate towards the positive plate. The potential difference between the plates is 12 V and the electron travels a distance of 0.15 m (0.10 m + 0.050 m) while gaining potential energy. Thus, the gain in potential energy is:
ΔPE = qΔV = (1.6 × 10^-19 C)(12 V) = 1.92 × 10^-18 J
This gain in potential energy is equal to the loss in kinetic energy of the electron. At a distance of 0.050 m from the positive plate, the electron has lost all of its initial potential energy and has gained an equal amount of kinetic energy. Thus, the kinetic energy gained by the electron is:
ΔKE = ΔPE = 1.92 × 10^-18 J
Converting to scientific notation in the required format gives:
ΔKE = 4.8 × 10^-19 J, which is choice B.

Question 10

In a region where the electric field is uniform and points in the +x direction,the electric potential is -2000 V at x = 8 m and is +400 V at x = 2 m.What is the magnitude of the electric field?

Accepted Answer

The electric field is given by the change in potential over the change in distance:
E = ΔV/Δx

ΔV = (+400 V) - (-2000 V) = 2400 V
Δx = (2 m) - (8 m) = -6 m (magnitude)

Note that the magnitude of Δx is used, since we are only interested in the magnitude of the electric field.

Therefore, the electric field is:
E = (2400 V) / (-6 m) = -400 V/m

However, since the electric field is given to be in the +x direction, we take the magnitude:
|E| = 400 V/m

Question 11

An electron is released from rest at a distance of 9.00 cm from a fixed proton.How fast will the electron be moving when it is 3.00 cm from the proton?,(k = 1/4πε0 = 8.99 × 109 N ∙ m²/C², ,

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Question 12

The figure shows an arrangement of two particles each having charge Q = -6.8 nC and each separated by 5.0 mm from a proton.If the two particles are held fixed at their locations and the proton is set into motion as shown,what is the minimum speed the proton needs to totally escape from these particles? (m_proton = 1.67 × 10^-27 kg,e = 1.60 × 10^-19 C,k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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Question 13

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across its plates is 100 V?

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Question 14

Consider a uniform horizontal electric field of 50 N/C directed toward the east.If the electric potential at a given point in the field is 80 V,what is the potential at a point 1.0 m directly south of that point?

Accepted Answer

The electric potential does not change when moving perpendicular to the direction of the electric field. Since the field is horizontal and directed east, moving south does not change the potential.

Question 15

A space probe approaches a planet,taking measurements as it goes.If it detects a potential difference of 6000 MV between the altitudes of 253,000 km and 276,000 km above the planet's surface,what is the approximate electric field strength produced by the planet at 264,500 km above the surface? Assume the electric field strength is approximately constant at these altitudes.

Accepted Answer

The electric field strength (E) can be calculated using the formula E = V/d, where V is the potential difference and d is the distance between the two points. Here, V = 6000 MV = 6 x 10^9 V and d = 276,000 km - 253,000 km = 23,000 km = 23 x 10^6 m. Thus, E = (6 x 10^9 V) / (23 x 10^6 m) ≈ 261 N/C.

Question 16

A battery maintains the electrical potential difference of 6.0-V between two large parallel metal plates separated by 1.0 mm.What is the strength of the electric field between the plates?

Accepted Answer

The strength of the electric field (E) between two plates is given by the formula E = V/d, where V is the potential difference and d is the separation between the plates. Here, V = 6.0 V and d = 1.0 mm = 0.001 m. Thus, E = 6.0 V / 0.001 m = 6000 V/m.

Question 17

The figure shows a group of three particles,all of which have charge Q = 8.8 nC.How much work did it take to assemble this group of charges if they all started out extremely far from each other? (k = 1/4πε₀

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Question 18

Each plate of an ideal air-filled parallel-plate capacitor has an area of 0.0010 ,and the separation of the plates is An electric field of Is present between the plates.What is the capacitance of this capacitor? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Accepted Answer

The answer of Each plate of an ideal air-filled parallel-plate...

Question 19

Consider a uniform horizontal electric field of 50 N/C directed toward the east.If the electric potential measured at a given point is 80 V,what is the potential at a point 1.0 m directly west of that point?

Accepted Answer

The potential difference (ΔV) in an electric field (E) over a distance (d) is given by ΔV = Ed. Here, E = 50 N/C, and d = 1.0 m (west means against the field direction). Thus, ΔV = 50 N/C * 1.0 m = 50 V. Since moving against the field direction increases potential, the potential 1.0 m west is 80 V + 50 V = 130 V.

Question 20

A uniform electric field,with a magnitude of 500 V/m,is points in the +x direction.If the potential at 
Is 2500 V,what is the potential at 
?

Accepted Answer

The answer of A uniform electric field,with a magnitude of...

Quiz 20: Electric Potential and Electric

Each plate of an ideal air-filled parallel-plate capacitor has an area of 0.0020 ,and the separation of the plates is An electric field of Is present between the plates.What is the surface charge density on the plates? (ε0 = 8.85 × 10-12 C2/N ∙ m2)

An ideal air-filled parallel-plate capacitor with horizontal plates has a plate separation of 5.0 cm.If the potential difference between the plates is 2000 V,with the top plate at the higher potential,what are the magnitude and direction of the electric field between the plates?

Two isolated copper plates,each of area 0.40 m2,carry opposite charges of magnitude 7.08 × 10-10 C.They are placed opposite each other in parallel alignment,with a spacing of 4.0 cm between them.What is the potential difference between the plates? (ε0 = 8.85 × 10-12 C2/N ∙ m2)

A proton moves 0.10 m along the direction of an electric field of magnitude 3.0 V/m.What is the change in kinetic energy of the proton? (e = 1.60 × 10-19C)

An ideal air-filled parallel plate capacitor with plate a separation of 4.0 cm has a plate area of 0.040 m2.What is the capacitance of this capacitor with air between these plates? (ε0 = 8.85 × 10-12 C2/N ∙ m2)