Question 1

A 1.0 μF capacitor has a potential difference of 6.0 V applied across its plates.If the potential difference across its plates is increased to 8.0 V,how much additional energy does the capacitor store?

Accepted Answer

A

Question 2

Two parallel plates are separated by 1.0 mm.If the potential difference between them is 3.0 V,what is the magnitude of their surface charge densities?

Accepted Answer

The surface charge density is given by the formula:$$\sigma = \frac{Q}{A}$$where Q is the charge on the plate and A is the area of the plate.In this problem, the potential difference between the plates is 3.0 V and the distance between the plates is 1.0 mm. The area of the plates is not given, but we can assume that it is large enough that the edge effects can be neglected.The charge on the plates can be found using the formula:$$Q = CV$$where C is the capacitance of the plates and V is the potential difference between the plates.The capacitance of the plates can be found using the formula:$$C = \frac{\epsilon_0 A}{d}$$where $\epsilon_0$ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.Substituting the given values into the formulas above, we get:$$Q = CV = \frac{\epsilon_0 A}{d}V$$$$\sigma = \frac{Q}{A} = \frac{\epsilon_0 V}{d}$$$$\sigma = \frac{(8.85 	imes 10^{-12} 	ext{ F/m})(3.0 	ext{ V})}{1.0 	imes 10^{-3} 	ext{ m}} = 27 	ext{ nC/m}^2$$Therefore, the magnitude of the surface charge densities is 27 nC/m^2.

Question 3

A proton with speed 1.5 × 10⁵ m/s falls through a potential difference of 100 volts,gaining speed.What is the speed reached?

Accepted Answer

When a proton falls through a potential difference (ΔV), it gains kinetic energy (K) given as K=qΔV (where q is the charge of the proton). Using this equation, we get K = (1.602 × 10^-19 C)(100 V) = 1.602 × 10^-17 J. The gained kinetic energy is equal to the change in kinetic energy, which is given as ΔK = 0.5mv^2 - 0.5mv0^2 (where m is the mass of the proton, v0 is the initial speed of the proton, and v is the final speed of the proton). Rearranging this equation, we get v = sqrt((2K/m) + v0^2). Plugging in the given values, we get v = sqrt((2(1.602 × 10^-17 J)/(1.6726 × 10^-27 kg)) + (1.5 × 10^6 m/s)^2) ≈ 2.04 × 10^6 m/s. Therefore, the answer is B.

Question 4

Two parallel circular plates with radius 7.0 mm carrying equal-magnitude surface charge densities of ±3.0 μC/m² are separated by a distance of 1.0 mm.How much stored energy do the plates have?

Accepted Answer

The stored energy is given by U = (1/2)CV^2, where C is the capacitance of the plates and V is the potential difference between them. The capacitance of two parallel plates is given by C = εA/d, where ε is the permittivity of the vacuum, A is the area of the plates, and d is the distance between them. The potential difference between the plates is given by V = Ed, where E is the electric field between the plates. The electric field between two parallel plates is given by E = σ/ε, where σ is the surface charge density. Substituting these expressions into the equation for stored energy, we get:U = (1/2)(εA/d)(Ed)^2 = (1/2)εA(σ^2/ε) = (1/2)σ^2APlugging in the given values, we get:U = (1/2)(3.0 μC/m^2)^2(π(7.0 mm)^2) = 78 nJ

Question 5

A 6.0 μC negative charge is attracted to a large,well-anchored,positive charge.How much kinetic energy does the negatively charged object gain if the potential difference through which it moves is 3 mV?

Accepted Answer

The kinetic energy gained by the charge is equal to the work done on it, which is the product of the charge and the potential difference. KE = qV = (6.0 x 10^-6 C)(3 x 10^-3 V) = 18 x 10^-9 J = 18 nJ.

Question 6

Two parallel plates that are initially uncharged are separated by 1.2 mm.What charge must be transferred from one plate to the other if 11.0 kJ of energy are to be stored in the plates? The area of each plate is 19.0 mm².

Accepted Answer

The answer of Two parallel plates that are initially uncharged...

Question 7

A charge of 3.0 μC and a second charge are initially far apart.If it takes 29 J of work to bring them to a final configuration in which the 3.0 μC is at x = 1.0 mm,y - 1.0 mm,and the other charge is at x = 1.0 mm,y = 3.0 mm,find the magnitude of the unknown charge.(The value of k is 9.0 $\times$ 10S1U1P19S1S1P0 N·m²/C². )

Accepted Answer

The work done in bringing the charges together is equal to the potential energy of the final configuration. The potential energy of two charges q1 and q2 separated by a distance r is given by:$$U = k * q1 * q2 / r$$where k is the Coulomb constant.In this problem, q1 = 3.0 μC, q2 is the unknown charge, and r is the distance between the charges. We can use the given information to find the value of q2.$$U = 29 J$$$$r = \sqrt{(1.0 mm)^2 + (3.0 mm)^2} = 3.16 mm$$$$k = 9.0 * 10^9 N \cdot m^2 / C^2$$Substituting these values into the equation for U, we get:$$29 J = (9.0 * 10^9 N \cdot m^2 / C^2) * (3.0 \mu C) * q2 / (3.16 mm)$$Solving for q2, we get:$$q2 = 4.30 \mu C$$Therefore, the magnitude of the unknown charge is 4.30 μC.

Question 8

Consider the group of charges in this figure.All three charges have Q = 2.4 nC.What is their electric potential energy?

Accepted Answer

The answer of Consider the group of charges in this...

Question 9

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of ±1.0 nC/mm².If the potential between the plates is 23.0 kV,find the surface area of one of the plates.

Accepted Answer

The capacitance C is given by C = ε₀(A/d), where A is the area and d is the separation. The charge Q is given by Q = σA, where σ is the surface charge density. Using Q = CV, we find A = C*V/σ. Substituting the given values, A = (3.0 pF * 23.0 kV) / (1.0 nC/mm²) = 69 mm².

Question 10

A sphere with radius 2.0 mm carries a 1.0 μC charge.What is the potential difference,VB - VA,between point B 4.0 m from the center of the sphere and point A 9.0 m from the center of the sphere? (The value of k is 9.0 $\times$ 10S1U1P19S1S1P0 N·m²/C². )

Accepted Answer

The potential difference is calculated using the formula V = kQ(1/rA - 1/rB). Substituting the given values, VB - VA = 9.0 x 10^9 x 1.0 x 10^-6 (1/9 - 1/4) = 1300 V.

Question 11

A 100.0 pF capacitor consists of two circular plates of radius 0.40 mm.How far apart are the plates? (The value of εo is 8.85 $\times$ 10S1U1P1-12S1S1P0 C²/N·m². )

Accepted Answer

The capacitance of a parallel-plate capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$where:* C is the capacitance in farads (F)* $\epsilon_0$ is the permittivity of free space (8.85 x 10^-12 C^2/N·m^2)* A is the area of the plates in square meters (m^2)* d is the distance between the plates in meters (m)We are given the following information:* C = 100.0 pF = 100.0 x 10^-12 F* r = 0.40 mm = 0.40 x 10^-3 m* $\epsilon_0$ = 8.85 x 10^-12 C^2/N·m^2We can use the formula above to solve for d:$$d = \frac{\epsilon_0 A}{C}$$$$d = \frac{(8.85 x 10^-12 C^2/N·m^2)(\pi (0.40 x 10^-3 m)^2)}{100.0 x 10^-12 F}$$$$d = 0.044 \mu m$$Therefore, the correct answer is A) 0.044 μm.

Question 12

An electron was accelerated from rest through a potential difference of 9900 V.What is its speed?

Accepted Answer

The kinetic energy gained by the electron when it's accelerated through a potential difference is given by

KE = qV

Where q is the charge on the electron and V is the potential difference.

Substituting the given values, we get

KE = (1.6 × 10^-19 C) × 9900 V = 1.584 × 10^-15 J

The kinetic energy can also be expressed as

KE = (1/2)mv^2

Where m is the mass of the electron and v is its final speed.

Substituting the known values, we get

v = √(2KE/m) = √(2 × 1.584 × 10^-15 J/9.11 × 10^-31 kg) = 5.9 × 10^7 m/s

Therefore, the correct answer is (A) 5.9 × 10^7 m/s.

Question 13

A 3.0 μC point charge and a 9.0 μC point charge are initially infinitely far apart.How much work does it take to bring the 3.0 μC point charge to x = 3.0 mm,y = 0.0 mm,and the 9.0 μC point charge to x = -3.0 mm,y = 0.0 mm? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

Accepted Answer

The work done in bringing the charges from infinity to the given positions is given by:$$W = k\frac{q_1q_2}{r_{12}}$$where k is Coulomb's constant, q1 and q2 are the charges, and r12 is the distance between the charges.In this case, q1 = 3.0 μC, q2 = 9.0 μC, and r12 = 6.0 mm = 0.006 m.Substituting these values into the equation, we get:$$W = (9.0 	imes 10^9 	ext{ N$\cdot$m}^2/	ext{C}^2)\frac{(3.0 	imes 10^{-6} 	ext{ C})(9.0 	imes 10^{-6} 	ext{ C})}{0.006 	ext{ m}}$$$$W = 41 	ext{ J}$$Therefore, the work done in bringing the charges from infinity to the given positions is 41 J.

Question 14

A parallel-plate capacitor has a voltage of 391 V applied across its plates,then the voltage source is removed.What is the voltage across its plates if the space between them becomes filled with mica? (The dielectric constant for mica is 5.4 and for air is 1.0006. )

Accepted Answer

When the dielectric is inserted, the voltage across the plates decreases by a factor equal to the dielectric constant. The new voltage is 391 V / 5.4 = 72.41 V, approximately 72 V.

Question 15

A dipole with ±6.0 μC charges is positioned so that the positive charge is 1.0 mm to the right of the origin and the negative charge is at the origin.How much work does it take to bring a 10.0 μC charge from infinity to a position x = 3.0 mm,y = 0.0 mm? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

Accepted Answer

The work done in bringing a charge q from infinity to a point near a dipole is given by $$W=k\frac{qQ}{r^2}$$where k is the Coulomb constant, Q is the charge of the dipole, q is the charge brought from infinity and r is the distance between the charge and the dipole.In this case, Q = 6.0 μC, q = 10.0 μC, and r = 3.0 mm = 0.003 m. Substituting these values into the equation, we get:$$W=9.0	imes10^9\frac{(6.0	imes10^{-6})(10.0	imes10^{-6})}{(0.003)^2}=90	ext{ J}$$

Question 16

A parallel-plate capacitor has a potential energy due to its charge of 6.00 mJ.It is accidentally filled with water in such a way as not to discharge its plates.How much energy does it continue to store after it is filled? (The dielectric constant for water is 78 and for air is 1.0006. )

Accepted Answer

The energy stored in a capacitor is inversely proportional to the dielectric constant. When the capacitor is filled with water, the dielectric constant increases from 1.0006 to 78, reducing the energy stored by a factor of 78. Therefore, the new energy is 6.00 mJ / 78 ≈ 0.077 mJ.

Question 17

Point charges,Q₁ = +46 nC and Q₂ = -75 nC,are placed as shown.In the figure,an external force transports an electron from point A to point B.The work done by the external force is closest to:

Accepted Answer

The answer of Point charges,Q₁ = +46 nC and...

Question 18

A parallel-plate capacitor consists of two parallel,square plates that have dimensions 1.0 cm by 1.0 cm.If the plates are separated by 1.0 mm,and the space between them is filled with teflon,what is the capacitance? (The dielectric constant for teflon is 2.1. )

Accepted Answer

The capacitance of a parallel-plate capacitor is given by C = (ε₀ * κ * A) / d, where ε₀ is the permittivity of free space (8.85 x 10⁻¹² F/m), κ is the dielectric constant, A is the area of the plates, and d is the separation between the plates. Substituting the given values, C = (8.85 x 10⁻¹² F/m * 2.1 * 1.0 x 10⁻⁴ m²) / 1.0 x 10⁻³ m = 1.9 x 10⁻¹² F = 1.9 pF.

Question 19

Three charges form an equilateral triangle with 1.6 cm long sides.What is the electric potential at the point indicated with the dot?

Accepted Answer

The answer of Three charges form an equilateral triangle with...

Question 20

A proton with a speed of 2.0 x 10⁵ m/s falls through a potential difference V and thereby increases its speed to 4.0 × 10⁵ m/s.Through what potential difference did the proton fall?

Accepted Answer

The change in kinetic energy of the proton is equal to the work done by the electric field, which is the charge of the proton times the potential difference. Using the formula ΔKE = qV, where ΔKE = 0.5 * m * (v_f^2 - v_i^2), and knowing the charge of a proton is approximately 1.6 x 10^-19 C, we can solve for V to find it is approximately 630 V.

Quiz 21: Electric Potential

A 1.0 μF capacitor has a potential difference of 6.0 V applied across its plates.If the potential difference across its plates is increased to 8.0 V,how much additional energy does the capacitor store?

Two parallel plates are separated by 1.0 mm.If the potential difference between them is 3.0 V,what is the magnitude of their surface charge densities?

A proton with speed 1.5 × 10⁵ m/s falls through a potential difference of 100 volts,gaining speed.What is the speed reached?

Two parallel circular plates with radius 7.0 mm carrying equal-magnitude surface charge densities of ±3.0 μC/m² are separated by a distance of 1.0 mm.How much stored energy do the plates have?

A 6.0 μC negative charge is attracted to a large,well-anchored,positive charge.How much kinetic energy does the negatively charged object gain if the potential difference through which it moves is 3 mV?

Two parallel plates that are initially uncharged are separated by 1.2 mm.What charge must be transferred from one plate to the other if 11.0 kJ of energy are to be stored in the plates? The area of each plate is 19.0 mm².

Consider the group of charges in this figure.All three charges have Q = 2.4 nC.What is their electric potential energy?

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of ±1.0 nC/mm².If the potential between the plates is 23.0 kV,find the surface area of one of the plates.

A sphere with radius 2.0 mm carries a 1.0 μC charge.What is the potential difference,VB - VA,between point B 4.0 m from the center of the sphere and point A 9.0 m from the center of the sphere? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

A 100.0 pF capacitor consists of two circular plates of radius 0.40 mm.How far apart are the plates? (The value of εo is 8.85 $\times$ 10^-12 C²/N·m². )

An electron was accelerated from rest through a potential difference of 9900 V.What is its speed?

A 3.0 μC point charge and a 9.0 μC point charge are initially infinitely far apart.How much work does it take to bring the 3.0 μC point charge to x = 3.0 mm,y = 0.0 mm,and the 9.0 μC point charge to x = -3.0 mm,y = 0.0 mm? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

A parallel-plate capacitor has a voltage of 391 V applied across its plates,then the voltage source is removed.What is the voltage across its plates if the space between them becomes filled with mica? (The dielectric constant for mica is 5.4 and for air is 1.0006. )

A parallel-plate capacitor has a potential energy due to its charge of 6.00 mJ.It is accidentally filled with water in such a way as not to discharge its plates.How much energy does it continue to store after it is filled? (The dielectric constant for water is 78 and for air is 1.0006. )

Point charges,Q₁ = +46 nC and Q₂ = -75 nC,are placed as shown.In the figure,an external force transports an electron from point A to point B.The work done by the external force is closest to:

A parallel-plate capacitor consists of two parallel,square plates that have dimensions 1.0 cm by 1.0 cm.If the plates are separated by 1.0 mm,and the space between them is filled with teflon,what is the capacitance? (The dielectric constant for teflon is 2.1. )

Three charges form an equilateral triangle with 1.6 cm long sides.What is the electric potential at the point indicated with the dot?

A proton with a speed of 2.0 x 10⁵ m/s falls through a potential difference V and thereby increases its speed to 4.0 × 10⁵ m/s.Through what potential difference did the proton fall?

Quiz 21: Electric Potential

A 1.0 μF capacitor has a potential difference of 6.0 V applied across its plates.If the potential difference across its plates is increased to 8.0 V,how much additional energy does the capacitor store?

Two parallel plates are separated by 1.0 mm.If the potential difference between them is 3.0 V,what is the magnitude of their surface charge densities?

A proton with speed 1.5 × 105 m/s falls through a potential difference of 100 volts,gaining speed.What is the speed reached?

Two parallel circular plates with radius 7.0 mm carrying equal-magnitude surface charge densities of ±3.0 μC/m2 are separated by a distance of 1.0 mm.How much stored energy do the plates have?

A 6.0 μC negative charge is attracted to a large,well-anchored,positive charge.How much kinetic energy does the negatively charged object gain if the potential difference through which it moves is 3 mV?

Two parallel plates that are initially uncharged are separated by 1.2 mm.What charge must be transferred from one plate to the other if 11.0 kJ of energy are to be stored in the plates? The area of each plate is 19.0 mm2.

Consider the group of charges in this figure.All three charges have Q = 2.4 nC.What is their electric potential energy?

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of ±1.0 nC/mm2.If the potential between the plates is 23.0 kV,find the surface area of one of the plates.

A sphere with radius 2.0 mm carries a 1.0 μC charge.What is the potential difference,VB - VA,between point B 4.0 m from the center of the sphere and point A 9.0 m from the center of the sphere? (The value of k is 9.0 ×\times× 109 N·m2/C2. )

A 100.0 pF capacitor consists of two circular plates of radius 0.40 mm.How far apart are the plates? (The value of εo is 8.85 ×\times× 10-12 C2/N·m2. )

An electron was accelerated from rest through a potential difference of 9900 V.What is its speed?

A 3.0 μC point charge and a 9.0 μC point charge are initially infinitely far apart.How much work does it take to bring the 3.0 μC point charge to x = 3.0 mm,y = 0.0 mm,and the 9.0 μC point charge to x = -3.0 mm,y = 0.0 mm? (The value of k is 9.0 ×\times× 109 N·m2/C2. )

A parallel-plate capacitor has a voltage of 391 V applied across its plates,then the voltage source is removed.What is the voltage across its plates if the space between them becomes filled with mica? (The dielectric constant for mica is 5.4 and for air is 1.0006. )

A parallel-plate capacitor has a potential energy due to its charge of 6.00 mJ.It is accidentally filled with water in such a way as not to discharge its plates.How much energy does it continue to store after it is filled? (The dielectric constant for water is 78 and for air is 1.0006. )

Point charges,Q1 = +46 nC and Q2 = -75 nC,are placed as shown.In the figure,an external force transports an electron from point A to point B.The work done by the external force is closest to:

A parallel-plate capacitor consists of two parallel,square plates that have dimensions 1.0 cm by 1.0 cm.If the plates are separated by 1.0 mm,and the space between them is filled with teflon,what is the capacitance? (The dielectric constant for teflon is 2.1. )

Three charges form an equilateral triangle with 1.6 cm long sides.What is the electric potential at the point indicated with the dot?

A proton with a speed of 2.0 x 105 m/s falls through a potential difference V and thereby increases its speed to 4.0 × 105 m/s.Through what potential difference did the proton fall?

A proton with speed 1.5 × 10⁵ m/s falls through a potential difference of 100 volts,gaining speed.What is the speed reached?

Two parallel circular plates with radius 7.0 mm carrying equal-magnitude surface charge densities of ±3.0 μC/m² are separated by a distance of 1.0 mm.How much stored energy do the plates have?

Two parallel plates that are initially uncharged are separated by 1.2 mm.What charge must be transferred from one plate to the other if 11.0 kJ of energy are to be stored in the plates? The area of each plate is 19.0 mm².

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of ±1.0 nC/mm².If the potential between the plates is 23.0 kV,find the surface area of one of the plates.

A sphere with radius 2.0 mm carries a 1.0 μC charge.What is the potential difference,VB - VA,between point B 4.0 m from the center of the sphere and point A 9.0 m from the center of the sphere? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

A 100.0 pF capacitor consists of two circular plates of radius 0.40 mm.How far apart are the plates? (The value of εo is 8.85 $\times$ 10^-12 C²/N·m². )

A 3.0 μC point charge and a 9.0 μC point charge are initially infinitely far apart.How much work does it take to bring the 3.0 μC point charge to x = 3.0 mm,y = 0.0 mm,and the 9.0 μC point charge to x = -3.0 mm,y = 0.0 mm? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

Point charges,Q₁ = +46 nC and Q₂ = -75 nC,are placed as shown.In the figure,an external force transports an electron from point A to point B.The work done by the external force is closest to:

A proton with a speed of 2.0 x 10⁵ m/s falls through a potential difference V and thereby increases its speed to 4.0 × 10⁵ m/s.Through what potential difference did the proton fall?