Question 1

A 6.9 ?C 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 = 1/4??₀ = 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:$$V = k\frac{q}{r} = (9.0 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2)\frac{(-6.9 	imes 10^{-6} 	ext{ C})}{0.016 	ext{ m}} = -12 	ext{ MV}$$The negative sign indicates that the potential difference is in the opposite direction of the electric field.

Question 2

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 = 1/4??₀ = 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 sphere, and r is the distance from the point to the center of the sphere. Substituting these expressions into the equation for the work done, we get:$$W = q\left(k\frac{q}{r_B} - k\frac{q}{r_A}\right)$$$$W = k\frac{q^2}{r_B} - k\frac{q^2}{r_A}$$$$W = k\frac{q^2}{r_A}\left(\frac{r_A}{r_B} - 1\right)$$$$W = k\frac{q^2}{r_A}\left(\frac{6.0 \mathrm{~m}}{4.0 \mathrm{~m}} - 1\right)$$$$W = k\frac{q^2}{r_A}\left(\frac{2}{4} - 1\right)$$$$W = k\frac{q^2}{r_A}\left(-\frac{1}{2}\right)$$$$W = -\frac{1}{2}k\frac{q^2}{r_A}$$The potential difference between points B and A is then given by:$$V_B - V_A = -\frac{W}{q}$$$$V_B - V_A = -\left(-\frac{1}{2}k\frac{q^2}{r_A}\right)$$$$V_B - V_A = \frac{1}{2}k\frac{q^2}{r_A}$$$$V_B - V_A = \frac{1}{2}\left(9.0 \times 10^9 \mathrm{~N \cdot m^2/C^2}\right)\frac{(2.0 \times 10^{-6} \mathrm{~C})^2}{(6.0 \mathrm{~m})}$$$$V_B - V_A = 1500 \mathrm{~V}$$Therefore, the potential difference between point B and point A is 1500 V.

Question 3

Two 5.0-µC point charges are 12 cm apart. What is the electric potential (relative to infinity)of this combination at the point where the electric field due to these charges is zero? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

To find the point where the electric field due to two point charges is zero, we use the fact that at this point the electric potential due to the two charges will be equal in magnitude and opposite in sign. Let the distance of this point from each of the charges be x. Then we can use the equation for electric potential due to a point charge:

V=kQ/x

The electric potential at this point due to the first charge is:

V1=kQ1/x

The electric potential at this point due to the second charge is:

V2=kQ2/(12-x)

Since the electric field due to the two charges is zero at this point, we have:

V1=-V2

Substituting the expressions for V1 and V2 and solving for x, we get:

x=4.8 cm

The total electric potential at this point is then:

V=kQ1/x+kQ2/(12-x)

Substituting the values, we get:

V=1.5 MV

Therefore, the answer is B.

Question 4

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 answer of A proton that is initially at rest...

Question 5

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 6

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

First, we need to calculate the electric potential (V) at the midpoint of each side of the rectangle due to each point charge using the formula:

V = kq/d

where k is Coulomb's constant, q is the charge of the point charge, and d is the distance between the point charge and the midpoint of the side.

Calculating the electric potential at the midpoint of the longer side of the rectangle:

V1 = (9.0 × 10^9 N ∙ m^2/C^2)(2.0 × 10^-6 C)/(2.0 cm) = 9.0 V

Calculating the electric potential at the midpoint of one of the shorter sides of the rectangle:

V2 = (9.0 × 10^9 N ∙ m^2/C^2)(2.0 × 10^-6 C)/(1.5 cm) = 12.0 V

Since these two sides are parallel to each other and have the same electric potential at their midpoints, we can use the equation for the electric potential due to parallel plates to find the electric potential at the center of the rectangle:

Vcenter = (V1 + V2)/2

Vcenter = (9.0 V + 12.0 V)/2 = 10.5 V

Finally, we convert this electric potential to megavolts:

Vcenter = 10.5 V = 10.5 × 10^-6 MV ≈ 2.9 MV

Therefore, the best choice is B.

Question 7

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 8

Two 3.0 ?C 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 = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

Accepted Answer

The potential at a point due to a point charge is given by $ V = \frac{kQ}{r} $. Calculate the potential due to each charge at 6.0 m and sum them: $ V = \frac{9.0 \times 10^9 \times 3.0 \times 10^{-6}}{6} + \frac{9.0 \times 10^9 \times 3.0 \times 10^{-6}}{4} = 4500 + 6750 = 11250 \approx 11,000 $ V.

Question 9

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 10

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 answer of If an electron is accelerated from rest...

Question 11

A +5.0-µC point charge is 12 cm from a -5.0-µC point charge. What is the magnitude of the electric field they produce at the point on the line connecting them where their electric potential (relative to infinity)is zero? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of A +5.0-µC point charge is 12 cm...

Question 12

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

The kinetic energy gained by the proton is equal to the work done by the electric field. We can use the formula:
K = qV
where K is the kinetic energy gained, q is the charge of the proton (1.60 x 10^-19 C), and V is the potential difference (500 V). 
Plugging in the values, we get:
K = (1.60 x 10^-19 C)(500 V) = 8.0 x 10^-17 J
The kinetic energy gained is equal to 1/2 mv^2, where m is the mass of the proton (1.67 x 10^-27 kg) and v is the final velocity. Solving for v, we get:
v = sqrt((2K)/m) = sqrt((2(8.0 x 10^-17 J))/(1.67 x 10^-27 kg)) = 3.1 x 10^5 m/s or 3.1 x 10^15 nm/s
Therefore, the answer is B: 3.1 x 10^15 nm/s.

Question 13

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 equals the work done by the electric field: $ \Delta KE = q \Delta V $. The change in kinetic energy is $ \frac{1}{2} m (v_f^2 - v_i^2) $. Solving for $ \Delta V $, we get $ \Delta V = \frac{m (v_f^2 - v_i^2)}{2q} $. Substituting the given values, $ \Delta V = \frac{1.67 \times 10^{-27} \times ((4.0 \times 10^5)^2 - (2.0 \times 10^5)^2)}{2 \times 1.60 \times 10^{-19}} \approx 630 $ V.

Question 14

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 change in potential energy of the proton is given by:
ΔPE = qΔV
where q is the charge of the proton (which is the same as the charge of an electron, e = 1.60 × 10^-19 C) and ΔV is the potential difference through which it is accelerated.

ΔPE = (1.60 × 10^-19 C)(0.100 kV) = 1.60 × 10^-20 J

This change in potential energy is equal to the kinetic energy gained by the proton:

ΔPE = KEf - KEi

where KEi is the initial kinetic energy of the proton (which is given by (1/2)mv^2) and KEf is its final kinetic energy after acceleration.

(1/2)mv^2f = (1/2)mv^2i + ΔPE

Since the mass of a proton (mp) is given, we can solve for the final speed (vf):

vf = √[(2/mp) * [(1.60 × 10^-20 J) + (1/2)mp(1.50 × 10^5 m/s)^2]]

vf = 2.04 × 10^5 m/s

Therefore, the final speed of the proton is 2.04 × 10^5 m/s, which is answer B.

Question 15

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 needed to move a charge q through a potential difference V is given by the equation W=qV. In this case, we are moving a single electron (q=-e) from the positive terminal to the negative terminal of a 9.0-V battery, so the work needed is:
W=(-e)(9.0 V)=(-1.60 × 10^-19 C)(9.0 V)= -14.4 × 10^-19 J
Note that the negative sign indicates that work is done by the battery on the electron (i.e. the battery is the source of energy for the electron's movement). The mass of the electron is not needed to solve this problem. Therefore, the best choice is D.

Question 16

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 17

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 18

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 19

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 μC and -3.0 μC. What is the potential (relative to infinity)at the midpoint of the square? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

To find the potential at the midpoint of the square, we need to find the contributions of each charge and add them up. We can use the formula for the potential due to a point charge:

V = kq/r

where V is the potential, k is Coulomb's constant (k = 1/4πε0), q is the charge, and r is the distance from the charge to the point where we want to find the potential.

We can also use the principle of superposition, which says that the total potential due to a collection of charges is the sum of the potentials due to each charge.

Let's start with the charge of +4.0 μC in one corner. The distance from this charge to the midpoint of the square is d = s/2√2, which is the diagonal length of half the square. So we have:

V1 = (9.0 × 10^9 N ∙ m^2/C^2) (4.0 × 10^-6 C) / (s/2√2)
V1 = 2.25 × 10^4 V

The potential due to the charge of +3.0 μC in the opposite corner is the same, because the distances are the same:

V2 = (9.0 × 10^9 N ∙ m^2/C^2) (3.0 × 10^-6 C) / (s/2√2)
V2 = 1.69 × 10^4 V

The potential due to the charge of -3.0 μC in the third corner is a little trickier, because this charge is negative. This means that the potential it creates is negative, and it also means that the direction of the electric field it creates is opposite to the direction created by the other two charges. To take this into account, we need to use the magnitude of the charge and the distance, but include a negative sign:

V3 = -(9.0 × 10^9 N ∙ m^2/C^2) (3.0 × 10^-6 C) / (s/2√2)
V3 = -1.69 × 10^4 V

Finally, we can add up the potentials to get the total potential at the midpoint:

Vtot = V1 + V2 + V3
Vtot = 2.25 × 10^4 V + 1.69 × 10^4 V - 1.69 × 10^4 V
Vtot = 3.25 × 10^4 V

However, the question asks for the potential relative to infinity. This means that we need to subtract the potential at infinity, which is defined to be zero:

V = Vtot - 0
V = 3.25 × 10^4 V - 0
V = 3.25 × 10^4 V

Therefore, the answer is B, 1.0 × 10^15 V.

Question 20

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.

Quiz 21: Electric Potential

Two 5.0-µC point charges are 12 cm apart. What is the electric potential (relative to infinity) of this combination at the point where the electric field due to these charges is zero? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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

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)

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)

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?

Two 3.0 ?C 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 = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

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?

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)

A +5.0-µC point charge is 12 cm from a -5.0-µC point charge. What is the magnitude of the electric field they produce at the point on the line connecting them where their electric potential (relative to infinity) is zero? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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)

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)

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)

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)

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)

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)

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 square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 μC and -3.0 μC. What is the potential (relative to infinity) at the midpoint of the square? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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?

Quiz 21: Electric Potential

Two 5.0-µC point charges are 12 cm apart. What is the electric potential (relative to infinity) of this combination at the point where the electric field due to these charges is zero? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

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

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)

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)

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?

Two 3.0 ?C charges lie on the x-axis, one at the origin and the other at 2.0 m2.0 \mathrm {~m}2.0 m What is the potential (relative to infinity) due to these charges at a point at 6.0 m6.0 \mathrm {~m}6.0 m on the x-axis? (k = 1/4??0 = 9.0 × 109 N ? m2/C2)

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?

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

A +5.0-µC point charge is 12 cm from a -5.0-µC point charge. What is the magnitude of the electric field they produce at the point on the line connecting them where their electric potential (relative to infinity) is zero? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

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 proton with a speed of 2.0 x 10510 ^ { 5 }105 m/s accelerates through a potential difference and thereby increases its speed to 4.0 x 10510 ^ { 5 }105 m/s. Through what magnitude potential difference did the proton accelerate? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)

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 × 10-19 C , mproton = 1.67 × 10-27 kg)

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)

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)

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)

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 square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 μC and -3.0 μC. What is the potential (relative to infinity) at the midpoint of the square? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

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?

Two 5.0-µC point charges are 12 cm apart. What is the electric potential (relative to infinity) of this combination at the point where the electric field due to these charges is zero? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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

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)

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)

Two 3.0 ?C 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 = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

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)

A +5.0-µC point charge is 12 cm from a -5.0-µC point charge. What is the magnitude of the electric field they produce at the point on the line connecting them where their electric potential (relative to infinity) is zero? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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)

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)

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)

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)

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)

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)

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 square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 μC and -3.0 μC. What is the potential (relative to infinity) at the midpoint of the square? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)