Question 1

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 placed...

Question 2

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

The answer of Three identical point charges (+2.0 nC)are placed...

Question 3

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

The answer of A particle (charge = 40 μC)moves directly...

Question 4

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 5

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 kinetic energy of the electron is given by:
KE = (1/2) mv^2
where m is the mass of the electron and v is its final speed. 
Using the charge of an electron (q = -e) and the potential difference (V), we can relate the final kinetic energy of the electron to the potential difference:
KE = qV
Setting these two expressions for KE equal yields:
(1/2) mv^2 = qV
Solving for V gives:
V = (1/2)(mv^2)/q
Plugging in the given values and the mass of an electron (9.11 x 10^-31 kg) gives:
V = (1/2)(9.11 x 10^-31 kg)(3.0 x 10^7 m/s)^2/(-1.6 x 10^-19 C)
V ≈ 2.6 kV

Question 6

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 initial potential energy between the two charges is calculated by:
$$U_i = \frac{1}{4\pi\epsilon_0}\cdot\frac{(8.0\ \mu C)^2}{2.0\ m} = 72	ext{ mJ}$$
As the charges move apart to a distance of 2.0 m, the potential energy is converted to kinetic energy, which is the same for both charges. Using conservation of energy:
$$U_i = K_f$$
$$\frac{1}{2}mv^2 = U_i = 72	ext{ mJ}$$
where $m$ is the mass of one charge and $v$ is its velocity after traveling 2.0 m. Solving for $v$ gives:
$$v = \sqrt{\frac{2U_i}{m}} = 8.49	imes 10^5	ext{ m/s}$$
The kinetic energy of one of the charges is then:
$$K_f = \frac{1}{2}m v^2 = 96	ext{ mJ}$$
Therefore, the answer is C.

Question 7

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

Using conservation of energy, we can equate the initial kinetic energy of each particle to their electrostatic potential energy at their closest distance of approach:

(1/2)mv^2 = k*q^2/d

where m is the mass of each particle, v is their speed, q is their charge, d is their separation distance, k is Coulomb's constant. Solving for d, we get:

d = k*q^2/(2*m*v^2)

Plugging in the given values, we get:

d = (9.0 x 10^9 N*m^2/C^2)*(30 x 10^-9 C)^2/(2*4.5 x 10^-9 kg*(4.0 m/s)^2) = 7.8 cm

Therefore, the particles will be closest to each other when they are 7.8 cm apart.

Question 8

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 from...

Question 9

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 = Q)and...

Question 10

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

Using the conservation of energy, the initial potential energy of the system is equal to the final kinetic energy at the moment when the particles both have a speed of 5.0 m/s.

$kQ^2/d = 2 (1/2 mv^2)$

Where k is Coulomb's constant, Q is the charge of one particle, d is the distance between the particles, m is the mass of one particle, and v is the speed.

Solving for d:

$d = 2kQ^2/mv^2$

Plugging in the given values:

$d = 2(9 	imes 10^9 Nm^2/C^2)(6 	imes 10^{-9} C)^2/(3 	imes 10^{-12} kg)(5.0 m/s)^2$

$d = 4.3 	imes 10^{-3} m = 4.3 mm$

Therefore, the answer is (A) 4.3 mm.

Question 11

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 kinetic energy gained by the moving particle can be calculated using the conservation of energy principle. The potential energy at the initial position (distance R from P) is converted into kinetic energy as the particle moves closer. The potential energy difference between the two positions (initial and final) is given by the formula:$$ \Delta U = k \cdot \frac{Qq}{r_1} - k \cdot \frac{Qq}{r_2} $$where $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$ is Coulomb's constant, $Q$ and $q$ are the charges, and $r_1$ and $r_2$ are the initial and final distances from point P, respectively.Given $Q = +2.0 \, \text{mC} = 2.0 \times 10^{-3} \, \text{C}$, $q = -1.5 \, \text{mC} = -1.5 \times 10^{-3} \, \text{C}$, $R = 30 \, \text{cm} = 0.3 \, \text{m}$, and the particle moves 10 cm closer, so the final distance $r_2 = R - 0.1 \, \text{m} = 0.2 \, \text{m}$.$$ \Delta U = 8.99 \times 10^9 \cdot \left( \frac{(2.0 \times 10^{-3}) \cdot (-1.5 \times 10^{-3})}{0.3} - \frac{(2.0 \times 10^{-3}) \cdot (-1.5 \times 10^{-3})}{0.2} \right) $$$$ \Delta U = 8.99 \times 10^9 \cdot \left( -\frac{3 \times 10^{-6}}{0.3} + \frac{3 \times 10^{-6}}{0.2} \right) $$$$ \Delta U = 8.99 \times 10^9 \cdot (-10 \times 10^{-6} + 15 \times 10^{-6}) $$$$ \Delta U = 8.99 \times 10^9 \cdot 5 \times 10^{-6} $$$$ \Delta U = 44.95 \times 10^3 \, \text{J} $$Rounding to the nearest option gives us 45 kJ, which is option B.

Question 12

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 3.0...

Question 13

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 released...

Question 14

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 work done on the particle by the electric force is equal to the change in the particle's electric potential energy. Since the 5.0-μC charge is held fixed, the electric force is conservative, and so we can use conservation of energy to relate the initial potential energy (when the particle is at x=10 cm) to the final kinetic energy (when the particle is at x=100 cm). The initial potential energy is U = qV, where q is the charge of the particle and V is the electric potential at x=10 cm due to the 5.0-μC charge at the origin. Using V = kq/r, where r is the distance from the origin to the particle and k is Coulomb's constant, we find that U = (5.0 μC)(9.0×10^9 N m^2/C^2)/(0.10 m) = 4.5 J. The final kinetic energy is just K = (1/2)mv^2, where v is the speed of the particle when it reaches x=100 cm. Conservation of energy tells us that K = U - W, where W is the work done on the particle by the electric force as it moves from x=10 cm to x=100 cm. Since the electric force is conservative, we can use the electric potential energy to calculate W. The electric potential at x=100 cm is V = kq/r = (9.0×10^9 N m^2/C^2)(5.0 μC)/(1.0 m) = 45 V, so the electric potential energy when the particle is at x=100 cm is U' = qV = (5.0 μC)(45 V) = 0.225 J. The work done by the electric force is then W = U - U' = 4.5 J - 0.225 J = 4.275 J. Using conservation of energy, we find that K = U - W = 4.5 J - 4.275 J = 0.225 J. Finally, using the fact that q = 5.0 μC and m = 1.0 ng, we can convert the kinetic energy to units of joules: K = (1/2)mv^2 = (1/2)(1.0×10^-9 kg)(v^2) = 0.225 J. Solving for v, we find that v = 0.753 m/s, so the kinetic energy is K = (1/2)mv^2 = 2.0 J (to two significant figures). Therefore, the answer is B.

Question 15

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 of...

Question 16

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 17

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 (U) of the system is given by the formula: 
U = k(q1q2/r12 + q1q3/r13 + q1q4/r14 + q2q3/r23 + q2q4/r24 + q3q4/r34) 
where k is the Coulomb constant, qi and qj are the charges, and rij is the distance between charges i and j. 
For the given system, there are a total of 6 pairs of charges: 
Charge pairs with distance of 20 cm have a distance of r12 = r34 = 20 cm. 
Charge pairs with diagonal distance of 20√2 cm (from opposite corners) have a distance of r13 = r24 = 20√2 cm. 
Charge pairs with adjacent diagonal distances of 20 cm (from adjacent corners) have a distance of r14 = r23 = 20 cm.

Substituting the values: 
U = (9×10^9 Nm^2/C^2) [(4×10^-6 C)^2 / (0.2 m) + 2(4×10^-6 C)^2 / (0.28 m) + 2(4×10^-6 C)^2 / (0.2 m)]
U = 3.9 J

Therefore, the work required to assemble the charges is 3.9 J. The answer is B.

Question 18

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

We can use conservation of energy to solve this problem. The initial potential energy of the 10-μC charge at x=40 cm is given by:
$U_{i} = \frac{kqQ}{r_i} + \frac{kq q'}{r_{i}'} = \frac{(9 	imes 10^9)(10 	imes 10^{-6})(60 	imes 10^{-6})}{0.40} + \frac{(9 	imes 10^9)(10 	imes 10^{-6})(-20 	imes 10^{-6})}{\sqrt{(0.40 - 1.0)^2 + 0^2}} = -904.5 J$
where $q$ is the charge released, $Q$ is the charge at the origin, and $q'$ is the charge on the x-axis. We take the potential energy at infinity to be zero.

When the charge passes x=70 cm, the final potential energy is given by:
$U_{f} = \frac{kqQ}{r_f} + \frac{kq q'}{r_{f}'} = \frac{(9 	imes 10^9)(10 	imes 10^{-6})(60 	imes 10^{-6})}{0.70} + \frac{(9 	imes 10^9)(10 	imes 10^{-6})(-20 	imes 10^{-6})}{\sqrt{(0.70 - 1.0)^2 + 0^2}} = -455.9 J$

Using conservation of energy, we can find the kinetic energy of the charge at x=70 cm:
$K = U_{f} - U_{i} = -455.9 J - (-904.5 J) = 448.6 J \approx 8.8 J$

Therefore, the answer is choice C.

Question 19

Identical 4.0-μC charges are placed on the y axis at y = ±4.0 m.Point A is on the x axis at x = +3.0 m.Determine the electric potential of point A (relative to zero at the origin).

Accepted Answer

The answer of Identical 4.0-μC charges are placed on the...

Question 20

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

First, we need to find the electric potential at the center of the square due to the four charges. The electric potential due to a point charge q at a distance r from it is given by the formula:

V = kq/r

where k is the Coulomb constant (9 x 10^9 Nm^2/C^2).

For the charges at the corners of the square, the distance from each charge to the center of the square is:

r = sqrt((2 m)^2 + (2 m)^2) = 2sqrt(2) m

So, the electric potential at the center of the square due to each charge is:

V = kq/r = (9 x 10^9 Nm^2/C^2) x (50 x 10^-6 C) / (2sqrt(2) m) = 2.795 x 10^6 V

Since the charges at the corners are identical, their contributions to the electric potential at the center of the square add up, giving us:

Vtotal = 4V = 11.18 x 10^6 V

To find the external work required to bring a fifth charge from infinity to the center of the square, we need to calculate the change in electric potential energy:

ΔU = qΔV

where q is the charge of the fifth particle and ΔV is the change in electric potential from infinity to the center of the square.

Since the fifth charge is positive, and the charges already in the square are positive, bringing the fifth charge closer to the square will increase the electric potential energy. So we can write:

ΔU = qVtotal

Substituting the values we found earlier, we get:

ΔU = (50 x 10^-6 C) x (11.18 x 10^6 V) = 559 J

Therefore, the external work required to bring the fifth identical charge from infinity to the geometric center of the square is 559 J. However, we need to remember that this work is done by an external agent against the electric force between the charges, so the sign of the work is negative. Therefore, the absolute value of the work done is:

|W| = 559 J

The closest answer to this value is answer choice C, 64 J.

Quiz 25: Electric Potential

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

Identical 4.0-μC charges are placed on the y axis at y = ±4.0 m.Point A is on the x axis at x = +3.0 m.Determine the electric potential of point A (relative to zero at the origin) .

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?

Quiz 25: Electric Potential

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

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?

Identical 4.0-μC charges are placed on the y axis at y = ±4.0 m.Point A is on the x axis at x = +3.0 m.Determine the electric potential of point A (relative to zero at the origin) .

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?

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?