Question 1

A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 2.0 m/s. What is the magnitude of the potential difference V?

Accepted Answer

We can use the formula for electrostatic potential energy to find the potential difference V:
U = qV
where U is the change in potential energy of the bead, q is the charge of the bead, and V is the potential difference.

We can also use the formula for kinetic energy to relate the change in potential energy to the final velocity of the bead:

U = (1/2)mv^2

Setting these two equations equal and solving for V, we get:

V = (1/2)mv^2/q

Substituting in the given values, we get:

V = (1/2)(0.004 kg)(2 m/s)^2/(20*10^-6 C) = 400 V

Therefore, the correct answer is D (400 V).

Question 2

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. How much kinetic energy does it gain? (e = 1.60 × 10^-19 C)

Accepted Answer

The kinetic energy gained by the proton is equal to the charge of the proton multiplied by the potential difference: KE = eV = (1.60 × 10^-19 C)(500 V) = 8.0 × 10^-17 J.

Question 3

How much work is needed to carry an electron from the positive terminal to the negative terminal of a 9.0-V battery. (e = 1.60 × 10^-19C , m_electron = 9.11 × 10^-31 kg)

Accepted Answer

The work required to move an electron from the positive terminal to the negative terminal of a battery is equal to the charge of the electron (1.60 × 10^-19 C) multiplied by the potential difference (9.0 V). Thus, the work is (1.60 × 10^-19 C) × (9.0 V) = 14.4 × 10^-19 J, which is closest to option D.

Question 4

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10^-19C , m_proton = 1.67 × 10^-27 kg)

Accepted Answer

We can use the equation: 
KE = (1/2)mv^2 = qV 
where KE is the kinetic energy gained by the proton, q is its charge, V is the potential difference, m is the mass of the proton, and v is its final velocity. 
Solving for v, we get: 
v = sqrt((2qV)/m) 
Plugging in the given values, we get: 
v = sqrt((2*1.60x10^-19 C * 500 V)/(1.67x10^-27 kg)) = 3.1x10^5 m/s 
Therefore, the answer is B.

Question 5

A tiny particle with charge + 5.0 μC is initially moving at 55 m/s. It is then accelerated through a potential difference of 500 V. How much kinetic energy does this particle gain during the period of acceleration?

Accepted Answer

The kinetic energy gained by the particle is equal to the work done on it, which can be calculated using the formula $W = qV$, where $q$ is the charge (in coulombs) and $V$ is the potential difference (in volts). Substituting the given values, $W = 5.0 \times 10^{-6} \, \text{C} \times 500 \, \text{V} = 2.5 \times 10^{-3} \, \text{J}$.

Question 6

How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The proton is 1836 times heavier than an electron, and e = 1.60 × 10^-19 C.

Accepted Answer

The kinetic energy gained by a charged particle when it is accelerated through a potential difference is given by the formula KE = qV, where q is the charge of the particle and V is the potential difference. In this case, the proton has a charge of +e and the potential difference is 1.00 V. Therefore, the kinetic energy gained by the proton is KE = (1.60 × 10^-19 C)(1.00 V) = 1.60 × 10^-19 J. Since the answer choices are all given in electron volts (eV) or joules (J), we need to convert the answer to either eV or J. To convert to eV, we divide by the electron volt to joule conversion factor, which is 1.60 × 10^-19 J/eV. Therefore, KE = (1.60 × 10^-19 J) / (1.60 × 10^-19 J/eV) = 1.00 eV, which is answer choice B.

Question 7

Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick. (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of Two very small +3.00-μC charges are at...

Question 8

A +4.0-μC and a -4.0-μC point charge are placed as shown in the figure. What is the potential difference between points A and B? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of A +4.0-μC and a -4.0-μC point charge...

Question 9

How much work must we do on an electron to move it from point A, which is at a potential of +50V, to point B, which is at a potential of -50 V, along the semicircular path shown in the figure? Assume the system is isolated from outside forces. (e = 1.60 × 10^-19C)

Accepted Answer

The work done on an electron is given by the equation W = -qΔV, where q is the charge of the electron and ΔV is the change in potential. Since the electron is moving from a positive potential to a negative potential, the change in potential is ΔV = (-50 V) - (+50 V) = -100 V. Therefore, the work done on the electron is

W = -(1.60 × 10^-19 C)(-100 V) 
= 1.60 × 10^-17 J

The answer is therefore choice B.

Question 10

A 6.9 μC negative point charge has a positively charged particle in an elliptical orbit about it. If the mass of the positively charged particle is $1.0 \mu \mathrm { g }$ and its distance from the point charge varies from $4.0 \mathrm {~mm}$ to $16.0 \mathrm {~mm}$ , what is the maximum potential difference through which the positive object moves? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The maximum potential difference is when the positive object is at its farthest point from the negative point charge. At this point, the potential difference is:$$V = k\frac{q}{r} = (9.0 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2)\frac{6.9 	imes 10^{-6} 	ext{ C}}{16.0 	imes 10^{-3} 	ext{ m}} = 12 	ext{ MV}$$

Question 11

If a Cu²⁺ ion that is initially at rest accelerates through a potential difference of 12 V without friction, how much kinetic energy will it gain? (e = 1.60 × 10^-19C)

Accepted Answer

The answer of If a Cu²⁺ ion that is initially...

Question 12

If it takes 50 J of energy to move 10 C of charge from point A to point B, what is the magnitude of the potential difference between points A and B?

Accepted Answer

The potential difference (voltage) between two points is defined as the work done per unit charge to move a charge between those two points. It can be calculated using the formula V = W/Q, where V is the potential difference, W is the work done or energy transferred, and Q is the charge. Substituting the given values, V = 50 J / 10 C = 5.0 V.

Question 13

If an electron is accelerated from rest through a potential difference of 1500 V, what speed does it reach? (e = 1.60 × 10^-19C , m_electron = 9.11 × 10^-31 kg)

Accepted Answer

The kinetic energy gained by the electron is equal to the work done on it by the electric field, which is eV. Using the equation $ \frac{1}{2}mv^2 = eV $, we can solve for v: $ v = \sqrt{\frac{2eV}{m}} $. Substituting the given values, $ v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 1500}{9.11 \times 10^{-31}}} \approx 2.3 \times 10^7 $ m/s.

Question 14

Three point charges, -2.00 μC, +4.00 μC, and +6.00 μC, are located along the x-axis as shown in the figure. What is the electric potential (relative to infinity) at point P due to these charges? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of Three point charges, -2.00 μC, +4.00 μC,...

Question 15

Four 2.0-µC point are at the corners of a rectangle with sides of length 3.0 cm and 4.0 cm. What is the electric potential (relative to infinity) at the midpoint of the rectangle? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

We can find the electric potential at the midpoint of the rectangle by summing up the potential contributions from each of the four point charges.

First, we need to find the distance from the midpoint of the rectangle to each of the four charges. Let the corners of the rectangle be labeled as A, B, C, and D, with AB = 4.0 cm and BC = 3.0 cm. Let the midpoint of the side AB be E, and the midpoint of the side BC be F. Then, the distance from E to each of the four charges is:

- AE = BE = CE = DE = sqrt((4.0/2)^2 + (3.0/2)^2) = sqrt(25/4) cm = 2.5 cm

Next, we can use the formula for the electric potential due to a point charge:

V = kq/r

where V is the potential, k is Coulomb's constant, q is the charge, and r is the distance from the charge.

Let Q be the magnitude of each charge (since they all have the same magnitude). Then, the potential due to each charge at the midpoint of the rectangle is:

- VA = VB = VC = VD = kQ/2.5 cm = (9.0 x 10^9 N*m^2/C^2)(2.0 x 10^-6 C)/(2.5 x 10^-2 m) = 7.2 x 10^5 V

To find the total potential at the midpoint, we need to add up the potentials due to all four charges:

Vtotal = VA + VB + VC + VD = 4(7.2 x 10^5 V) = 2.88 x 10^6 V

Since the question asks for the potential relative to infinity, we need to subtract off the potential at infinity, which is zero. Therefore, the final answer is:

Vmidpoint = Vtotal - Vinfinity = 2.88 x 10^6 V - 0 V = 2.88 x 10^6 V = 2.9 MV

Thus, the answer is B.

Question 16

At a distance d from a point charge Q, the energy density in its electric field is u. If we now go to a distance d/2 from the charge, what is the energy density at the new location?

Accepted Answer

The energy density of an electric field due to a point charge is proportional to the square of the electric field, which is inversely proportional to the square of the distance from the charge. Therefore, halving the distance increases the energy density by a factor of $ (d/d/2)^2 = 4^2 = 16 $.

Question 17

After a proton with an initial speed of 1.50 × 10⁵ m/s has increased its speed by accelerating through a potential difference of 0.100 kV, what is its final speed? (e = 1.60 × 10^-19C , m_proton = 1.67 × 10^-27 kg)

Accepted Answer

The kinetic energy gained by the proton is equal to the potential energy it lost while accelerating through the potential difference. Thus, we can use the following equation to find the final speed:

(1/2)mv^2 = qV

where m is the mass of the proton, v is its final speed, q is its charge, and V is the potential difference. Substituting the given values, we get:

(1/2)(1.67 × 10^-27 kg)(v^2) = (1.60 × 10^-19 C)(0.100 kV)

Solving for v, we get:

v = √((2 × 1.60 × 10^-19 C × 0.100 kV) / 1.67 × 10^-27 kg) = 2.04 × 10^5 m/s

Therefore, the best choice is B) 2.04 × 1015 m/s.

Question 18

A sphere with radius 2.0 mm carries a $+ 2.0 \mu \mathrm { C }$ charge. What is the potential difference, $V _ { B } - V _ { A }$ between point B, which is $4.0 \mathrm {~m}$ from the center of the sphere, and point A, which is $6.0 \mathrm {~m}$ from the center of the sphere? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The potential difference between two points in an electric field is given by the negative of the work done per unit charge in moving a charge from one point to the other. In this case, the work done in moving a charge from point A to point B is given by:$$W = q(V_B - V_A)$$where q is the charge of the sphere, and $V_B$ and $V_A$ are the potentials at points B and A, respectively.The potential at a point due to a point charge is given by:$$V = k\frac{q}{r}$$where k is Coulomb's constant, q is the charge of the point charge, and r is the distance from the point charge to the point where the potential is being calculated.In this case, the potential at point B is:$$V_B = k\frac{q}{r_B} = (9.0 \times 10^9 \mathrm{~N \cdot m^2/C^2})\frac{(2.0 \times 10^{-6} \mathrm{~C})}{4.0 \mathrm{~m}} = 4500 \mathrm{~V}$$and the potential at point A is:$$V_A = k\frac{q}{r_A} = (9.0 \times 10^9 \mathrm{~N \cdot m^2/C^2})\frac{(2.0 \times 10^{-6} \mathrm{~C})}{6.0 \mathrm{~m}} = 3000 \mathrm{~V}$$Therefore, the potential difference between point B and point A is:$$V_B - V_A = 4500 \mathrm{~V} - 3000 \mathrm{~V} = 1500 \mathrm{~V}$$

Question 19

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at $2.0 \mathrm {~m}$ What is the potential (relative to infinity) due to these charges at a point at $6.0 \mathrm {~m}$ on the x-axis? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The potential due to a point charge is given by:$$V = k\frac{q}{r}$$where:* k is Coulomb's constant (9.0 × 10^9 N ∙ m^2/C^2)* q is the charge (3.0 μC = 3.0 × 10^-6 C)* r is the distance from the charge to the point (6.0 m)The potential due to the charge at the origin is:$$V_1 = k\frac{q}{r} = (9.0 	imes 10^9 \frac{N \cdot m^2}{C^2})\frac{3.0 	imes 10^{-6} C}{6.0 m} = 4500 V$$The potential due to the charge at 2.0 m is:$$V_2 = k\frac{q}{r} = (9.0 	imes 10^9 \frac{N \cdot m^2}{C^2})\frac{3.0 	imes 10^{-6} C}{4.0 m} = 6750 V$$The total potential at 6.0 m is the sum of the potentials due to the two charges:$$V = V_1 + V_2 = 4500 V + 6750 V = 11,250 V$$Therefore, the correct answer is A) 11,000 V.

Question 20

A proton with a speed of 2.0 x $10^5$ m/s accelerates through a potential difference and thereby increases its speed to 4.0 x $10^5$ m/s. Through what magnitude potential difference did the proton accelerate? (e = 1.60 × 10^-19 C , m_proton = 1.67 × 10^-27 kg)

Accepted Answer

The change in kinetic energy of the proton is equal to the work done by the electric field, which is the charge times the potential difference. Using the formula: $ \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) = e \Delta V $, we find $ \Delta V = \frac{1}{2} \times 1.67 \times 10^{-27} \times ((4.0 \times 10^5)^2 - (2.0 \times 10^5)^2) / 1.60 \times 10^{-19} \approx 630 $ V.

Quiz 21: Electric Potential

A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 2.0 m/s. What is the magnitude of the potential difference V?

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. How much kinetic energy does it gain? (e = 1.60 × 10-19 C)

How much work is needed to carry an electron from the positive terminal to the negative terminal of a 9.0-V battery. (e = 1.60 × 10-19 C , melectron = 9.11 × 10-31 kg)

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)

A tiny particle with charge + 5.0 μC is initially moving at 55 m/s. It is then accelerated through a potential difference of 500 V. How much kinetic energy does this particle gain during the period of acceleration?

How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The proton is 1836 times heavier than an electron, and e = 1.60 × 10-19 C.

Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick. (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

A +4.0-μC and a -4.0-μC point charge are placed as shown in the figure. What is the potential difference between points A and B? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

How much work must we do on an electron to move it from point A, which is at a potential of +50V, to point B, which is at a potential of -50 V, along the semicircular path shown in the figure? Assume the system is isolated from outside forces. (e = 1.60 × 10-19C)

If a Cu2+ ion that is initially at rest accelerates through a potential difference of 12 V without friction, how much kinetic energy will it gain? (e = 1.60 × 10-19 C)