Question 1

Two capacitors with capacitances of 1.5 μC and 0.25 μF,respectively,are connected in parallel.The system is connected to a 40-V battery.What charge accumulates on the 1.5-μF capacitor?
A) 100 μC
B) 75 μC
C) 60 μC
D) 33 μC

Accepted Answer

The answer of Two capacitors with capacitances of 1.5 μC...

Question 2

Electrons in an x-ray machine are accelerated from rest through a potential difference of 80,000 V.What is the kinetic energy of each of these electrons in eV?
A) 128 eV
B) 40 eV
C) 80 eV
D) 80 keV

Accepted Answer

The kinetic energy gained by an electron accelerated through a potential difference of 80,000 V is equal to the charge of the electron (1 e) multiplied by the potential difference (80,000 V), which is 80,000 eV or 80 keV.

Question 3

Suppose two charges,one positive and one negative,are separated by a distance d.Where can two equipotential surfaces at different potentials intersect in the region near the charges?
A) halfway between the charges if the charges have the same magnitude
B) closer to the smaller negative charge than the larger positive charge if the charges are not equal
C) Equipotential surfaces at different potentials cannot intersect.
D) closer to the smaller positive charge than the larger negative charge if the charges are not equal

Accepted Answer

This is because equipotential surfaces are surfaces of constant potential, and if two different potentials intersect, it means there is a point on one surface that has two different potentials, which is not possible. The equipotential surfaces can only get closer together as they approach the charges, but they cannot intersect.

Question 4

If C = 48 µF,determine the equivalent capacitance for the combination shown. 
A) 32 µF
B) 48 µF
C) 24 µF
D) 28 µF

Accepted Answer

The answer of If C = 48 µF,determine the equivalent...

Question 5

Suppose two equipotential surfaces are the planes x = 3 m and x = 5 m.If the potential at the x = 3 m surface is 10 V and the potential at the x = 5 m surface is 50 V,what is the magnitude of the electric field between the two surfaces?
A) 40 N/C
B) 20 N/C
C) 15 N/C
D) 10 N/C

Accepted Answer

The potential difference between the two surfaces is 50 V - 10 V = 40 V. Since the potential difference is related to the electric field by ΔV = Ed, where d is the distance between the surfaces, we can solve for the electric field:
E = ΔV/d = (40 V)/(2 m) = 20 N/C. Therefore, the answer is choice B.

Question 6

If a doubly-ionized oxygen atom is accelerated from rest by going through a potential difference of 10 V,what will be the change in its kinetic energy?
A) 10 eV
B) 40 eV
C) 20 eV
D) none of the above

Accepted Answer

The change in kinetic energy of a charged particle accelerated through a potential difference can be calculated using the formula ΔK = qΔV, where q is the charge of the particle and ΔV is the potential difference. For a doubly-ionized oxygen atom, q = +2e, where e is the elementary charge. Thus, ΔK = 2e * 10 V = 20 eV. Therefore, the correct answer is (C) 20 eV.

Question 7

A 6.0-μF capacitor is attached to a 20-V power supply.How much energy is stored in the capacitor? A) 2.0 × 10−³ J B) 5.2 × 10−⁴ J C) 2.0 × 10−⁴ J D) 1.2 × 10−³ J

Accepted Answer

The energy stored in a capacitor is given by the formula $ E = \frac{1}{2} C V^2 $. Substituting the given values, $ C = 6.0 \times 10^{-6} \, \text{F} $ and $ V = 20 \, \text{V} $, we get $ E = \frac{1}{2} \times 6.0 \times 10^{-6} \times (20)^2 = 1.2 \times 10^{-3} \, \text{J} $.

Question 8

Two capacitors with capacitances of 1.5 μF and 0.25 μF,respectively,are connected in parallel.The system is connected to a 100-V battery.What electrical potential energy is stored in the 1.5-μF capacitor? A) 1.2 × 10−³ J B) 1.9 × 10−³ J C) 0.50 × 10−³ J D) 7.5 × 10−³ J

Accepted Answer

The answer of Two capacitors with capacitances of 1.5 μF...

Question 9

Two capacitors with capacitances of 1.0 μF and 0.50 μF,respectively,are connected in series.The system is connected to a 50-V battery.What electrical potential energy is stored in the 1.0-μF capacitor? A) 5.6 × 10−⁴ J B) 4.3 × 10−³ J C) 0.065 × 10−³ J D) 1.4 × 10−⁴ J

Accepted Answer

The potential difference across a capacitor in a series connection is given by V=Q/C, where V is the potential difference, Q is the charge stored on the capacitor, and C is the capacitance. Since both capacitors have the same charge, we can find the potential difference across each capacitor by using V=Q/C.

For the 1.0 μF capacitor, V = Q/1.0μF = Q*10^6 C/V

For the 0.50 μF capacitor, V = Q/0.50μF = 2Q*10^6 C/V

The total potential difference across the series connection is V_total = V_1 + V_2.

Using V_total = 50V, we can solve for Q, which is the same for both capacitors:

Q = CV = (1.0 μF + 0.50 μF) × (50 V) = 75 μC

The potential energy stored in a capacitor is given by U=1/2CV^2.

For the 1.0 μF capacitor, U = 1/2(1.0 μF)(25 V)^2 = 0.31 × 10^-6 J

Therefore, the answer is D: 1.4 × 10^-14 J (rounded to two significant figures).

Question 10

If three 3.0-µF capacitors are connected in parallel,what is the combined capacitance?
A) 0.46 μF
B) 1.0 μF
C) 9.0 μF
D) 0.75 μF

Accepted Answer

When capacitors are connected in parallel, their capacitance values add up. So, in this case, the combined capacitance is simply the sum of the individual capacitances:

C_total = C1 + C2 + C3
C_total = 3.0 µF + 3.0 µF + 3.0 µF
C_total = 9.0 µF

Therefore, the best choice is C, with a combined capacitance of 9.0 µF.

Question 11

An electron in a TV picture tube is accelerated through a potential difference of 20 kV before it hits the screen.What is the kinetic energy of the electron in electron volts? (1 eV = 1.6 × 10−¹⁹ J) A) 6.25 × 10²² eV B) 1.0 × 10⁴ eV C) 2.0 × 10⁴ eV D) 3.2 × 10−²² eV

Accepted Answer

The kinetic energy of an electron can be calculated using the formula:
Kinetic Energy = qV
where q is the charge of the electron and V is the potential difference.
The charge of an electron is -1.6 × 10^-19 C, and the potential difference in this case is 20 kV, which is equivalent to 20,000 V.
Therefore, the kinetic energy of the electron is:
Kinetic Energy = -1.6 × 10^-19 C × 20,000 V
Kinetic Energy = -3.2 × 10^-15 J
To convert this to electron volts (eV), we divide by the conversion factor of 1.6 × 10^-19 J/eV:
Kinetic Energy = (-3.2 × 10^-15 J) / (1.6 × 10^-19 J/eV)
Kinetic Energy = -20 eV
Since kinetic energy cannot be negative, we take the absolute value:
Kinetic Energy = 20 eV
Therefore, the correct answer is choice C: 2.0 × 10^4 eV.

Question 12

Two capacitors with capacitances of 1.0 μC and 0.50 μF,respectively,are connected in series.The system is connected to a 150 V battery.What charge accumulates on the 1.0-μF capacitor?
A) 50 μC
B) 150 μC
C) 100 μC
D) 33 μC

Accepted Answer

The answer of Two capacitors with capacitances of 1.0 μC...

Question 13

A 0.50-μF capacitor is connected to a 400-V battery.What potential energy is stored in the capacitor? A) 0.020 J B) 0.040 J C) 1.0 × 10−⁴ J D) 1.2 × 10−¹² J

Accepted Answer

The potential energy stored in a capacitor is given by the formula:

U = (1/2) * C * V^2

where U is the potential energy, C is the capacitance, and V is the voltage.

Substituting the given values, we get:

U = (1/2) * 0.50 * 10^-6 * (400)^2

U = 0.040 J

Therefore, the answer is B.

Question 14

A parallel-plate capacitor has a capacitance of 20 µF.What potential difference across the plates is required to store 22 × 10−⁵ C on this capacitor? A) 2.2 × 10^-2 V B) 1.4 × 10−⁸ V C) 11 V D) 70 V

Accepted Answer

The potential difference V is given by the formula V = Q/C, where Q is the charge and C is the capacitance. Substituting the given values, V = (22 × 10^-5 C) / (20 × 10^-6 F) = 11 V.

Question 15

A solid metal sphere of radius r is inside a metal concentric spherical shell of inner radius 2r and outer radius 3r.A charge is place on the solid metal sphere and it is found that the potential at its surface is V₁ (at r)and that the potential at the outer surface of the spherical shell is V₃ (at 3r).In terms of V₁ and V₃,what is the potential at the inner surface of the metal shell (at 2r)? A) 3V₁/2 B) V₃ C) V₁ D) 2V₃/2

Accepted Answer

The answer of A solid metal sphere of radius r...

Question 16

A 0.50-μF capacitor is connected to a 400 V battery.Find the charge on the capacitor. A) 2.0 × 10−⁴ C B) 0.020 C C) 1.0 × 10−⁴ C D) 1.2 × 10−¹² C

Accepted Answer

The charge on a capacitor is given by Q = C × V. Here, C = 0.50 μF = 0.50 × 10^-6 F and V = 400 V. So, Q = 0.50 × 10^-6 F × 400 V = 2.0 × 10^-4 C.

Question 17

If C = 14 µF,what is the equivalent capacitance for the combination shown? 
A) 5.8 µF
B) 6.5 µF
C) 8.2 µF
D) 7.0 µF

Accepted Answer

The answer of If C = 14 µF,what is the...

Question 18

What is the equivalent capacitance between points a and b? All capacitors are 2.0 μF. 
A) 1.2 µF
B) 0.25 µF
C) 4.0 µF
D) 0.60 µF

Accepted Answer

The answer of What is the equivalent capacitance between points...

Question 19

Equipotential surfaces around a point charge are
A) spherical.
B) cylindrical.
C) planar.
D) none of these choices.

Accepted Answer

The electric field lines around a point charge are radial, meaning they extend outward from the charge in all directions. Because the electric field is perpendicular to equipotential surfaces, the equipotential surfaces must also be radial and symmetric around the point charge, forming spherical surfaces. Therefore, choice A is the best answer.

Question 20

If the plane x = 4 m is an equipotential surface,in which of the following directions can the electric field be in the region immediately adjacent to this plane?
A) the positive y direction
B) both the positive and negative z directions
C) the positive x direction
D) more than one of these choices

Accepted Answer

An equipotential surface is a surface where the electric potential is constant. Therefore, the electric field at any point on this surface must be perpendicular to it. Since the surface in question is the plane x = 4 m, the only direction in which the electric field can be perpendicular to it is the positive x direction. Therefore, choice C is the correct answer.

Quiz 18: Direct-Current Circuits Sources of EMF: Part A

Two capacitors with capacitances of 1.5 μC and 0.25 μF,respectively,are connected in parallel.The system is connected to a 40-V battery.What charge accumulates on the 1.5-μF capacitor?

Electrons in an x-ray machine are accelerated from rest through a potential difference of 80,000 V.What is the kinetic energy of each of these electrons in eV?

Suppose two charges,one positive and one negative,are separated by a distance d.Where can two equipotential surfaces at different potentials intersect in the region near the charges?

If C = 48 µF,determine the equivalent capacitance for the combination shown.

Suppose two equipotential surfaces are the planes x = 3 m and x = 5 m.If the potential at the x = 3 m surface is 10 V and the potential at the x = 5 m surface is 50 V,what is the magnitude of the electric field between the two surfaces?

If a doubly-ionized oxygen atom is accelerated from rest by going through a potential difference of 10 V,what will be the change in its kinetic energy?

A 6.0-μF capacitor is attached to a 20-V power supply.How much energy is stored in the capacitor?

Two capacitors with capacitances of 1.5 μF and 0.25 μF,respectively,are connected in parallel.The system is connected to a 100-V battery.What electrical potential energy is stored in the 1.5-μF capacitor?

Two capacitors with capacitances of 1.0 μF and 0.50 μF,respectively,are connected in series.The system is connected to a 50-V battery.What electrical potential energy is stored in the 1.0-μF capacitor?

If three 3.0-µF capacitors are connected in parallel,what is the combined capacitance?

An electron in a TV picture tube is accelerated through a potential difference of 20 kV before it hits the screen.What is the kinetic energy of the electron in electron volts? (1 eV = 1.6 × 10−19 J)

Two capacitors with capacitances of 1.0 μC and 0.50 μF,respectively,are connected in series.The system is connected to a 150 V battery.What charge accumulates on the 1.0-μF capacitor?

A 0.50-μF capacitor is connected to a 400-V battery.What potential energy is stored in the capacitor?

A parallel-plate capacitor has a capacitance of 20 µF.What potential difference across the plates is required to store 22 × 10−5 C on this capacitor?

A 0.50-μF capacitor is connected to a 400 V battery.Find the charge on the capacitor.

If C = 14 µF,what is the equivalent capacitance for the combination shown?

What is the equivalent capacitance between points a and b? All capacitors are 2.0 μF.

Equipotential surfaces around a point charge are

If the plane x = 4 m is an equipotential surface,in which of the following directions can the electric field be in the region immediately adjacent to this plane?

An electron in a TV picture tube is accelerated through a potential difference of 20 kV before it hits the screen.What is the kinetic energy of the electron in electron volts? (1 eV = 1.6 × 10−¹⁹ J)

A parallel-plate capacitor has a capacitance of 20 µF.What potential difference across the plates is required to store 22 × 10−⁵ C on this capacitor?