Question 1

Four identical point charges (+4.0 μC) are placed at the corners of a square which has 20-cm sides. How much work is required to assemble this charge arrangement starting with each of the charges a very large distance from any of the other charges?

Accepted Answer

The potential energy of a system of four point charges is given by:
$U=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{12}}+\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{13}}+\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{14}}+\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{23}}+\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{24}}+\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{34}}$
where $r_{ij}$ is the distance between charges $i$ and $j$.

Since all charges are identical and equally spaced, we can set $r_{12}=r_{23}=r_{34}=a$ and $r_{13}=r_{24}=\sqrt{2}a$ where $a=20/2\sqrt{2}=10\sqrt{2}$ cm.

The total potential energy of the system is then:
$U=(6)\frac{1}{4\pi\epsilon_0}\frac{q^2}{a}+(4)\frac{1}{4\pi\epsilon_0}\frac{q^2}{\sqrt{2}a}$

To assemble the system, we need to bring each charge from infinity to its position in the square. The work done on each charge is equal to the change in potential energy, which is given by:

$W=\frac{1}{2}\Delta U$

We can find $\Delta U$ by subtracting the potential energy at infinity (which is zero) from the potential energy of the system:

$\Delta U=U-\frac{1}{4\pi\epsilon_0}\frac{q^2}{r_{\infty}}=U$

where $r_{\infty}$ is the distance from the charge to infinity.

Therefore, the work required to assemble the system is:

$W=(4)\frac{1}{2}U=2U$

Substituting the expression for $U$ and solving for $W$, we get:

$W=2(6+\sqrt{2})\frac{1}{4\pi\epsilon_0}\frac{(4.0	imes10^{-6}	ext{ C})^2}{10\sqrt{2}	ext{ cm}}$

Using the value for $\epsilon_0$, we get:

$W\approx3.876	ext{ J}$

The closest answer choice is B, 3.9 J.

Question 2

Two identical particles, each with a mass of 4.5 μg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?

Accepted Answer

The electric force between the two particles is attractive, so it will slow down their movement and eventually bring them to a stop before they move apart again. At this instant, they will be closest to each other. To find the distance between them at this point, we can use the conservation of energy:
Initial kinetic energy of both particles = final electric potential energy between them
(1/2)(2m)(v^2) = (kq^2)/r
where m is the mass of each particle, v is the speed of each particle, q is the charge of each particle, k is Coulomb's constant, and r is the distance between the particles.
Plugging in the given values and solving for r, we get:
r = (kq^2)/(2mv^2) = (9x10^9 Nm^2/C^2)(30x10^-9 C)^2/(2)(4.5x10^-6 kg)(4.0 m/s)^2
r = 0.078 m = 7.8 cm
Therefore, the particles will be closest to each other when they are 7.8 cm apart. The answer is (C).

Question 3

A particle (q = +5.0 μC) is released from rest when it is 2.0 m from a charged particle which is held at rest. After the positively charged particle has moved 1.0 m toward the fixed particle, it has a kinetic energy of 50 mJ. What is the charge on the fixed particle?

Accepted Answer

The answer of A particle (q = +5.0 μC) is...

Question 4

A particle (charge 7.5 μC) is released from rest at a point on the x axis, x = 10 cm. It begins to move due to the presence of a 2.0-μC charge which remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point x = 1.0 m?

Accepted Answer

The answer of A particle (charge 7.5 μC) is released...

Question 5

Two particles, each having a mass of 3.0 μg and having equal but opposite charges of magnitude 5.0 nC, are released simultaneously from rest when the two are 5.0 cm apart. What is the speed of either particle at the instant when the two are separated by 2.0 cm?

Accepted Answer

The answer of Two particles, each having a mass of...

Question 6

Identical point charges (+20 μC) are placed at the corners of an equilateral triangle with sides of 2.0-m length. How much external energy is required to bring a charge of 45 μC from infinity to the midpoint of one side of the triangle?

Accepted Answer

To solve this problem, we need to calculate the electric potential at the midpoint of one side of the triangle due to the charges at the corners, and then use that to calculate the external work required to bring the test charge from infinity to that point.

First, we need to calculate the electric potential at the midpoint of one side of the triangle. Since the charges at the corners are identical and equidistant from the midpoint, we can use the formula for the electric potential due to a point charge:

V = kq/r

where k is the Coulomb constant, q is the charge, and r is the distance from the charge to the point where we want to calculate the potential.

If we consider one of the charges at the corner of the triangle, the distance from that charge to the midpoint of one side is half the length of the side, or 1.0 m. Therefore, the electric potential due to that charge at the midpoint is:

V1 = kq/r = (9.0 x 10^9 N m^2/C^2)(20 x 10^-6 C)/(1.0 m) = 1.8 x 10^4 V

Since there are three identical charges at the corners, the total electric potential at the midpoint due to all three charges is three times this value:

Vtotal = 3V1 = 5.4 x 10^4 V

Now we can calculate the external work required to bring the test charge from infinity to the midpoint of one side of the triangle using the formula:

W = qΔV

where q is the test charge and ΔV is the change in electric potential. In this case, q = 45 x 10^-6 C and ΔV = Vtotal/2 (since we're going to the midpoint of one side, not the center of the triangle). Therefore:

W = (45 x 10^-6 C)(5.4 x 10^4 V/2) = 0.1215 J

Rounding to two significant figures, we get:

W = 0.12 J

which is closest to choice (D), 21 J.

Question 7

Identical point charges (+30 μC) are placed at the corners of a rectangle (4.0 m × 6.0 m). How much external energy is required to bring a charge of 55 μC from infinity to the midpoint of one of the 6.0-m long sides of the rectangle?

Accepted Answer

The answer of Identical point charges (+30 μC) are placed...

Question 8

Through what potential difference must an electron (starting from rest) be accelerated if it is to reach a speed of 3.0 × 107 m/s?

Accepted Answer

The answer of Through what potential difference must an electron...

Question 9

Four identical point charges (+6.0 nC) are placed at the corners of a rectangle which measures 6.0 m × 8.0 m. If the electric potential is taken to be zero at infinity, what is the potential at the geometric center of this rectangle?

Accepted Answer

The answer of Four identical point charges (+6.0 nC) are...

Question 10

A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC charge is released from rest at a point x = 40 cm, what is its kinetic energy the instant it passes the point x = 70 cm?

Accepted Answer

The answer of A 60-μC charge is held fixed at...

Question 11

Two particles, each having a mass of 3.0 μg and having equal but opposite charges of magnitude of 6.0 nC, are released simultaneously from rest when they are a very large distance apart. What distance separates the two at the instant when each has a speed of 5.0 m/s?

Accepted Answer

The energy conservation principle can be applied here, where the initial potential energy due to the electric force between the charges is converted into kinetic energy as they move towards each other. The initial potential energy (U) when the particles are infinitely apart is considered to be zero, and the final potential energy when they are at a distance $r$ apart is given by $U = k \frac{q_1 q_2}{r}$, where $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2$), and $q_1$ and $q_2$ are the charges of the particles. The kinetic energy (K) of each particle when they have a speed of $v = 5.0 \, \text{m/s}$ is given by $K = \frac{1}{2}mv^2$, where $m$ is the mass of the particle.Since the total energy is conserved, the sum of the kinetic energies of both particles equals the potential energy at the final position:$$2 \times \frac{1}{2}mv^2 = k \frac{q^2}{r}$$Given that $m = 3.0 \, \mu\text{g} = 3.0 \times 10^{-9} \, \text{kg}$, $q = 6.0 \, \text{nC} = 6.0 \times 10^{-9} \, \text{C}$, and $v = 5.0 \, \text{m/s}$, we can solve for $r$:$$mv^2 = k \frac{q^2}{r}$$$$r = k \frac{q^2}{mv^2}$$Plugging in the values:$$r = \frac{8.99 \times 10^9 \times (6.0 \times 10^{-9})^2}{3.0 \times 10^{-9} \times (5.0)^2}$$$$r \approx 4.3 \, \text{mm}$$Therefore, the correct answer is 4.3 mm, which corresponds to choice A.

Question 12

Identical point charges (+50 μC) are placed at the corners of a square with sides of 2.0-m length. How much external energy is required to bring a fifth identical charge from infinity to the geometric center of the square?

Accepted Answer

Since the charges are positive, the potential energy will increase as the fifth charge is brought closer. Using the formula for electric potential energy, we can calculate the energy required to bring the fifth charge from infinity to the center of the square:

$U=\frac{1}{4\pi\epsilon_0}\cdot\frac{q^2}{r}$

where q is the charge and r is the distance between the charges. Since the square has sides of 2.0 m, the distance from one corner to the center is $\sqrt{2}$ m. The distance from the fifth charge to the center is also $\sqrt{2}$ m, since the square is symmetrical. Therefore, the total distance the fifth charge needs to be moved is $\sqrt{2}+\sqrt{2}=2\sqrt{2}$ m.

Substituting in the values, we get:

$U=\frac{1}{4\pi\epsilon_0}\cdot\frac{(50\cdot10^{-6})^2}{2\sqrt{2}}=64$ J

Therefore, the external energy required is 64 J.

Question 13

A particle (charge = Q) is kept in a fixed position at point P, and a second particle (charge = q) is released from rest when it is a distance R from P. If Q = +2.0 mC, q = −1.5 mC, and R = 30 cm, what is the kinetic energy of the moving particle after it has moved a distance of 10 cm?

Accepted Answer

The potential energy gained by the moving particle due to the electrostatic interaction with the fixed particle at point P can be calculated using the formula:

$U = \frac{k|qQ|}{r}$

where k is Coulomb's constant, q and Q are the charges of the moving and fixed particles respectively, and r is the distance between them.

Initial potential energy of the system when the moving particle is at a distance R = 30 cm from P:

$U_i = \frac{k|qQ|}{R} = \frac{(9 	imes 10^9 Nm^2/C^2) 	imes (1.5 	imes 10^{-3} C) 	imes (2 	imes 10^{-3} C)}{0.3 m} = 90 J$

When the moving particle travels a distance of 10 cm, its distance from P decreases to r = R - 10 cm = 20 cm. Therefore, the final potential energy of the system is:

$U_f = \frac{k|qQ|}{r} = \frac{(9 	imes 10^9 Nm^2/C^2) 	imes (1.5 	imes 10^{-3} C) 	imes (2 	imes 10^{-3} C)}{0.2 m} = 135 J$

The difference between the final and initial potential energies is equal to the kinetic energy gained by the moving particle:

$K = U_f - U_i = 135 J - 90 J = 45 J = 45 kJ$

Therefore, the kinetic energy of the moving particle after it has moved a distance of 10 cm is 45 kJ (option B).

Question 14

A particle (charge = 40 μC) moves directly toward a second particle (charge = 80 μC) which is held in a fixed position. At an instant when the distance between the two particles is 2.0 m, the kinetic energy of the moving particle is 16 J. Determine the distance separating the two particles when the moving particle is momentarily stopped.

Accepted Answer

Using conservation of energy, we can find the velocity of the moving particle at the point where it stops: 
$$
\frac{1}{2}mv^2 = 16 J \implies v = 4\frac{	ext{m}}{	ext{s}}
$$
At the moment when the moving particle stops, the electrostatic force between the two particles is equal and opposite to the force that was propelling the moving particle forward. Thus, we can use Coulomb's law to find the distance between the particles at this point:
$$
k\frac{q_1q_2}{r^2} = ma \implies k\frac{q_1q_2}{r^2} = \frac{mv^2}{r} \implies r = \frac{kq_1q_2}{mv^2} = 0.95 	ext{ m}
$$
Thus, the answer is (C).

Question 15

A charge of +3.0 μC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energy is required to bring a charge of 25μC from infinity to the center of the circle?

Accepted Answer

The electric potential at the center of the circle due to the distribution of charge on the circumference is given by:
$$V=\frac{kQ}{r}$$
where k is the Coulomb constant, Q is the total charge distributed along the circumference and r is the radius of the circle.
We have Q = +3.0 μC and r = 20 cm = 0.2 m.
Therefore, 
$$V=\frac{(9	imes10^9\ Nm^2/C^2)(3.0	imes10^{-6}C)}{0.2\ m}=1.35	imes10^5\ V$$
Now, to bring a charge of 25μC from infinity to the center of the circle, we need to do work against the electric field. The work done in moving a charge q through a potential difference V is given by:
$$W=qV$$
Plugging in the values, we get:
$$W=(25	imes10^{-6}C)(1.35	imes10^5\ V)=3.38\ J$$
Therefore, the external energy required is 3.4 J (to two significant figures).

Question 16

Three identical point charges (+2.0 nC) are placed at the corners of an equilateral triangle with sides of 2.0-m length. If the electric potential is taken to be zero at infinity, what is the potential at the midpoint of any one of the sides of the triangle?

Accepted Answer

First, we need to find the distance between the midpoint of one side and the three charges. Let's consider the triangle formed by the midpoint and the two charges at the adjacent corners. This is a right triangle with legs of length 1m and hypotenuse of length $\sqrt{3}$ m (since it's an equilateral triangle with side length of 2m). We can use the Pythagorean theorem to find the distance between the midpoint and each adjacent charge:

$d = \sqrt{(\sqrt{3}/2)^2 + (1/2)^2} = \sqrt{3}/2$

Next, we need to calculate the potential at the midpoint due to each charge and then add them up. The potential due to a point charge is given by:

$V = k \frac{q}{r}$

where $k$ is the Coulomb constant ($9.0 	imes 10^9$ Nm$^2$/C$^2$), $q$ is the charge, and $r$ is the distance from the charge to the point where we are calculating the potential.

The potential at the midpoint due to each adjacent charge is:

$V = (9.0 	imes 10^9) \frac{2.0 	imes 10^{-9}}{\frac{\sqrt{3}}{2}} = 8.27 	ext{ V}$ (since $q$ is positive, the potential is positive)

Notice that these two potentials are identical due to the symmetry of the problem.

For the third charge, we need to find the distance from its center to the midpoint of the side, which is the same as the distance from the center to the midpoint of the equilateral triangle. This is just $2/\sqrt{3}$ m (half the height of the equilateral triangle). Thus, the potential at the midpoint due to the third charge is:

$V = (9.0 	imes 10^9) \frac{2.0 	imes 10^{-9}}{\frac{2}{\sqrt{3}}} = 3.27 	ext{ V}$

Adding up the potentials due to all three charges gives:

$V_{total} = 2(8.27 	ext{ V}) + 3.27 	ext{ V} = 19.81 \approx 20 	ext{ V}$

Note that the potential due to the charges at the adjacent corners will be the same, and it will be twice the potential due to the charge at the opposite corner, due to symmetry. Also, since the electric potential is a scalar quantity, we can add up the electric potentials due to each charge (by the principle of superposition), as we did above.

Question 17

Two identical particles, each with a mass of 2.0 μg and a charge of 25 nC, are released simultaneously from rest when the two are 4.0 cm apart. What is the speed of either particle at the instant when the two are separated by 10 cm?

Accepted Answer

The answer of Two identical particles, each with a mass...

Question 18

Identical 8.0-μC point charges are positioned on the x axis at x = ±1.0 m and released from rest simultaneously. What is the kinetic energy of either of the charges after it has moved 2.0 m?

Accepted Answer

The electric potential energy of the two charges when they are 2.0 m apart is:
$$U_{i} = \frac{kq^{2}}{r_{i}} = \frac{(9.0 	imes 10^{9} \mathrm{N \cdot m^{2}/C^{2}})(8.0 	imes 10^{-6} \mathrm{C})^{2}}{2.0 \mathrm{m}} = 2.88 \mathrm{J}$$
The total initial electric potential energy is $U_{total} = 2U_{i} = 5.76 \mathrm{J}$.
When the charges are 2.0 m apart, they have the same magnitude of charge and the same kinetic energy:
$$K = \frac{1}{2}mv^{2} = \frac{1}{2}\frac{U_{total} - U_{i}}{2} = 0.96 \mathrm{J} = 960 \mathrm{mJ}$$
Therefore, the kinetic energy of either charge is $960 \mathrm{mJ}$, which is closest to answer choice C, 96 mJ.

Question 19

Particle A (mass = m, charge = Q) and B (mass = m, charge = 5 Q) are released from rest with the distance between them equal to 1.0 m. If Q = 12 μC, what is the kinetic energy of particle B at the instant when the particles are 3.0 m apart?

Accepted Answer

The answer of Particle A (mass = m, charge =...

Question 20

A particle (charge = 5.0 μC) is released from rest at a point x = 10 cm. If a 5.0-μC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 90 cm?

Accepted Answer

The answer of A particle (charge = 5.0 μC) is...

Quiz 26: Electric Potential