Question 1

A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of −9.0 μC is at x = 1.0 m.At what point on the x axis is the electrical potential zero?
A) +0.33 m
B) −0.25 m
C) +0.25 m
D) +0.75 m

Accepted Answer

To find the point on the x-axis where the electrical potential is zero, we need to calculate the distance where the electric potential due to the positive charge is equal in magnitude but opposite in sign to the electric potential due to the negative charge. We can use the formula for electric potential due to a point charge:

V = kq/r

where k is the Coulomb constant (9 × 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.

At a distance x from the positive charge, the electric potential is:

V1 = kq1/x

At a distance 1-x from the negative charge, the electric potential is:

V2 = kq2/(1-x)

Setting V1 = -V2 and solving for x, we get:

kq1/x = -kq2/(1-x)

Simplifying and solving for x, we get:

x = (q1/q2) / (1 + q1/q2)

Plugging in the values, we get:

x = (3.0 × 10^-6 C) / (-9.0 × 10^-6 C) / (1 + 3.0/9.0)

x = 0.25 m

Therefore, the electrical potential is zero at a distance of 0.25 m from the positive charge, which corresponds to answer choice C.

Question 2

A proton (+1.6 × 10−¹⁹ C)moves 10 cm along the direction of an electric field of strength 4.0 N/C.The electrical potential difference between the proton's initial and ending points is: A) 4.8 × 10−¹⁹ V. B) 0.40 V. C) 0.044 V. D) 6.4 × 10−¹⁹ V.

Accepted Answer

The potential difference (V) is calculated using V = E × d, where E is the electric field strength and d is the distance. Here, V = 4.0 N/C × 0.10 m = 0.40 V.

Question 3

When charges q_a,q_b,and q_c are placed respectively at the corners a,b,and c of a right triangle,the potential at the midpoint of the hypotenuse is 20 V.When the charge q_a is removed,the potential at the midpoint becomes 15 V.When,instead,the charge q_b is removed (q_a and q_c both in place),the potential at the midpoint becomes 13 V.What is the potential at the midpoint if both charges q_a and q_c are removed? A) 13 V B) 7 V C) 5 V D) 8 V

Accepted Answer

When q_a is removed, the potential at the midpoint becomes 15V. This means that the potential due to q_a alone was 5V (20V-15V). When q_b is removed, the potential at the midpoint becomes 13V. This means that the potential due to q_b alone was 7V (20V-13V). If both q_a and q_c are removed, the potential at the midpoint will be the potential due to q_b alone, which is 7V. Thus, the answer is B) 7V.

Question 4

When charges q_a,q_b,and q_c are placed respectively at the corners a,b,and c of a right triangle,the potential at the midpoint of the hypotenuse is 20 V.When the charge q_a is removed,the potential at the midpoint becomes 15 V.When,instead,the charge q_b is removed (q_a and q_c both in place),the potential at the midpoint becomes 13 V.What is the potential at the midpoint if only the charge q_c is removed from the array of charges? A) 12 V B) 8 V C) 7 V D) 5 V

Accepted Answer

The potential at the midpoint of the hypotenuse is given by:$$V = \frac{1}{4\pi\epsilon_0} \left(\frac{q_a}{r_a} + \frac{q_b}{r_b} + \frac{q_c}{r_c}\right)$$where r_a, r_b, and r_c are the distances from the charges to the midpoint.When q_a is removed, the potential becomes:$$V = \frac{1}{4\pi\epsilon_0} \left(\frac{q_b}{r_b} + \frac{q_c}{r_c}\right)$$When q_b is removed, the potential becomes:$$V = \frac{1}{4\pi\epsilon_0} \left(\frac{q_a}{r_a} + \frac{q_c}{r_c}\right)$$When q_c is removed, the potential becomes:$$V = \frac{1}{4\pi\epsilon_0} \left(\frac{q_a}{r_a} + \frac{q_b}{r_b}\right)$$We can see that the potential when q_c is removed is the same as the potential when q_a is removed. Therefore, the potential at the midpoint when only q_c is removed is 15 V.

Question 5

An electron is released from rest at the negative plate of a parallel-plate capacitor.If the distance across the plates is 5.0 mm and the potential difference across the plates is 20 V,with what velocity does the electron hit the positive plate? (m_e = 9.1 × 10−³¹ kg,e = 1.6 × 10−¹⁹ C) A) 1.3 × 10⁶ m/s B) 1.0 × 10⁶ m/s C) 2.7 × 10⁶ m/s D) 5.3 × 10⁶ m/s

Accepted Answer

The answer of An electron is released from rest at...

Question 6

Two charges,each of charge q,are separated by a distance d.The distance between the charges is doubled.If the original potential energy of the arrangement was 40 mJ,by how much does the potential energy change when the charges are moved to the increased distance?
A) It decreases by more than 20 mJ.
B) It decreases by less than 20 mJ
C) More information is needed.
D) It decreases by 20 mJ.

Accepted Answer

The potential energy between two charges is inversely proportional to the distance between them. Doubling the distance halves the potential energy, so it decreases by 20 mJ (from 40 mJ to 20 mJ).

Question 7

Find the electrical potential at 0.055 m from a point charge of 6.0 μC.(k_e = 8.99 × 10⁹ N⋅m²/C²) A) 1.5 × 10⁵ V B) 1.2 × 10⁷ V C) 9.8 × 10⁵ V D) 3.6 × 10⁵ V

Accepted Answer

The answer of Find the electrical potential at 0.055 m...

Question 8

An ion is released from rest and moves due to the force from an electric field from a position in the field having a potential of 14 V to a position having a potential of 8 V.The ion:
A) can have either a positive or a negative charge.
B) must be neutral.
C) must have a positive charge.
D) must have a negative charge.

Accepted Answer

The ion moves from a higher potential (14 V) to a lower potential (8 V). This means that it is moving in the direction of the electric field, which is from higher potential to lower potential. Since the electric field is directed from positive to negative charge, the ion must have a positive charge in order to move in the direction of the electric field. Therefore, the correct answer is that the ion must have a positive charge (choice C).

Question 9

A solid conducting sphere of 10 cm radius has a net charge of 40 nC.If the potential at infinity is taken as zero,what is the potential at the center of the sphere? A) 360 μV B) 3.6 × 10³ V C) >3.6 × 10⁴ V D) 36 μV

Accepted Answer

The electric potential at the center of a charged sphere is given by V= kQ/R, where k is the Coulomb constant, Q is the charge on the sphere, and R is the radius of the sphere. Substituting the given values, we get V= (9 x 10^9 Nm^2/C^2) × (40 x 10^-9 C)/(0.1 m) = 3.6 x 10^5 V. However, the question asks for the potential with respect to infinity, so we need to subtract the potential at infinity, which is 0. Therefore, the potential at the center of the sphere is 3.6 × 10^5 V or 3.6 × 10^2 μV. The closest option is B, which is 3.6 × 10^2 μV. Option C is incorrect because it suggests that the potential is greater than 3.6 × 10^5 V, which is not possible for a single charged sphere.

Question 10

An electron in a cathode ray tube is accelerated through a potential difference of 10 kV.What kinetic energy does the electron gain in the process? (e = 1.6 × 10−¹⁹ C) A) 8.0 × 10−¹⁶ J B) 1.6 × 10−¹⁵ J C) 8.0 × 10²² J D) 1.6 × 10−¹⁶ J

Accepted Answer

The kinetic energy gained by the electron is equal to the charge of the electron multiplied by the potential difference: KE = eV = (1.6 × 10^-19 C)(10,000 V) = 1.6 × 10^-15 J.

Question 11

A proton (+1.6 × 10−¹⁹ C)moves 10 cm on a path in the direction opposite to a uniform electric field of strength 3.0 N/C.How much work is done on the proton by the electrical field? A) 4.8 × 10−²⁰ J B) 1.6 × 10−²⁰ J C) zero D) −4.8 × 10−²⁰ J

Accepted Answer

The work done by the electric field is given by the formula W = Fd, where W is the work done, F is the force, and d is the displacement. In this case, the force on the proton is given by F = qE, where q is the charge of the proton and E is the electric field strength. Since the proton moves in a direction opposite to the electric field, the force on the proton is also in the opposite direction, i.e., opposite to the displacement. Therefore, the work done by the electric field is negative.

Putting the values, we have: 
q = +1.6 × 10^-19 C (charge of proton)
E = 3.0 N/C (electric field strength)
d = -0.1 m (displacement)

Substituting in the formula, we get:
W = (1.6 × 10^-19 C) x (3.0 N/C) x (-0.1 m) = -4.8 × 10^-20 J

Therefore, the work done on the proton by the electric field is -4.8 × 10^-20 J or approximately -5 × 10^-20 J. The answer is D, -4.8 × 10^-20 J.

Question 12

Two particles each have the same mass but particle #1 has four times the charge of particle #2.Particle #1 is accelerated from rest through a potential difference of 10 V and attains speed v.Particle #2 is accelerated from rest also through a potential difference of 10 V.What speed does particle #2 attain?
A) 2v
B) v/2
C) v/4
D)

Accepted Answer

The answer of Two particles each have the same mass...

Question 13

A 9.0-V battery moves 200 mC of charge through a circuit running from its positive terminal to its negative terminal.How much energy was delivered to the circuit? A) 4.5 x 10³ J B) 1.8 J C) 2.2 mJ D) 0.18 J

Accepted Answer

The answer of A 9.0-V battery moves 200 mC of...

Question 14

At what distance from a point charge of 14 μC would the electrical potential be 4.2 × 10⁴ V? (k_e = 8.99 × 10⁹ N⋅m²/C²) A) 0.76 m B) 0.58 m C) 3.0 m D) 1.7 m

Accepted Answer

The answer of At what distance from a point charge...

Question 15

A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of −9.0 μC is at x = 1.00 m.What is the electric potential at the x = 0.50 m point? (k_e = 8.99 × 10⁹ N⋅m²/C²) A) −5.4 × 10⁴ V B) 11 × 10⁴ V C) 16 × 10⁴ V D) −11 × 10⁴ V

Accepted Answer

The electric potential at a point due to a point charge is given by V = k_e * q / r. Calculate the potential due to each charge at x = 0.50 m and sum them: V_total = V_1 + V_2 = (8.99 × 10^9 * 3.0 × 10^-6 / 0.50) + (8.99 × 10^9 * -9.0 × 10^-6 / 0.50) = 5.4 × 10^4 - 16.2 × 10^4 = -10.8 × 10^4 V, which is approximately -11 × 10^4 V.

Question 16

An electron (charge −1.6 × 10−¹⁹ C)moves on a path perpendicular to the direction of a uniform electric field of strength 3.0 N/C.How much work is done on the electron as it moves 20 cm? A) 1.6 × 10−²⁰ J B) −4.8 × 10−²⁰ J C) 4.8 × 10−²⁰ J D) zero

Accepted Answer

Since the electric field is perpendicular to the motion of the electron, the electric force does no work on the electron. Therefore, the work done on the electron is zero.

Question 17

If an electron is accelerated from rest through a potential difference of 300 V,find its approximate speed at the end of this process.(e = 1.6 × 10−¹⁹ C;m_e = 9.1 × 10−³¹ kg) A) 2.5 × 10⁷ m/s B) 2.1 × 10⁷ m/s C) 1.0 × 10⁷ m/s D) 1.4 × 10⁷ m/s

Accepted Answer

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

Question 18

A uniform electric field,with a magnitude of 600 N/C,is directed parallel to the positive x axis.If the potential at x = 4.0 m is 1000 V,what is the potential at x = 2.0 m?
A) −800 V
B) 2200 V
C) 2800 V
D) 1600 V

Accepted Answer

The potential difference in a uniform electric field is given by V = Ed, where E is the electric field and d is the distance. The potential decreases in the direction of the field. So, V(2.0 m) = V(4.0 m) + E*(4.0 m - 2.0 m) = 1000 V + 600 N/C * 2.0 m = 2200 V.

Question 19

Two protons,each of charge 1.60 × 10−¹⁹ C,are 3.00 × 10−⁵ m apart.What is the change in potential energy if they are brought 1.00 × 10−⁵ m closer together? (k_e = 8.99 × 10⁹ N·m²/C²) A) 3.84 × 10−²⁴ J B) 3.20 × 10−¹⁶ J C) 1.15 × 10−²³ J D) 3.20 × 10−¹⁹ J

Accepted Answer

The change in potential energy is given by:$$\Delta U = k_e \frac{q_1 q_2}{r_f} - k_e \frac{q_1 q_2}{r_i}$$where:* $k_e$ is the Coulomb constant* $q_1$ and $q_2$ are the charges of the protons* $r_f$ is the final distance between the protons* $r_i$ is the initial distance between the protonsPlugging in the given values, we get:$$\Delta U = (8.99 imes 10^9 ext{ N}\cdot ext{m}^2/ ext{C}^2) \frac{(1.60 imes 10^{-19} ext{ C})^2}{(3.00 imes 10^{-5} ext{ m})} - (8.99 imes 10^9 ext{ N}\cdot ext{m}^2/ ext{C}^2) \frac{(1.60 imes 10^{-19} ext{ C})^2}{(4.00 imes 10^{-5} ext{ m})}$$$$\Delta U = 3.84 imes 10^{-24} ext{ J}$$Therefore, the change in potential energy is 3.84 × 10−²⁴ J.

Question 20

A uniform electric field,with a magnitude of 600 N/C,is directed parallel to the positive x axis.If the potential at x = 4.0 m is 1000 V,what is the change in potential energy of a proton as it moves from x = 4.0 m to x = 1.0 m? (q_p = 1.6 × 10−¹⁹ C) A) 500 J B) 2.9 × 10−¹⁶ J C) 1.9 × 10−¹⁶ J D) 8.0 × 10−¹⁷ J

Accepted Answer

The answer of A uniform electric field,with a magnitude of...

Quiz 18: Direct-Current Circuits Sources of EMF: Part A

A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of −9.0 μC is at x = 1.0 m.At what point on the x axis is the electrical potential zero?

A proton (+1.6 × 10−¹⁹ C) moves 10 cm along the direction of an electric field of strength 4.0 N/C.The electrical potential difference between the proton's initial and ending points is:

An electron is released from rest at the negative plate of a parallel-plate capacitor.If the distance across the plates is 5.0 mm and the potential difference across the plates is 20 V,with what velocity does the electron hit the positive plate? (m_e = 9.1 × 10−³¹ kg,e = 1.6 × 10−¹⁹ C)

Two charges,each of charge q,are separated by a distance d.The distance between the charges is doubled.If the original potential energy of the arrangement was 40 mJ,by how much does the potential energy change when the charges are moved to the increased distance?

Find the electrical potential at 0.055 m from a point charge of 6.0 μC.(k_e = 8.99 × 10⁹ N⋅m²/C²)

An ion is released from rest and moves due to the force from an electric field from a position in the field having a potential of 14 V to a position having a potential of 8 V.The ion:

A solid conducting sphere of 10 cm radius has a net charge of 40 nC.If the potential at infinity is taken as zero,what is the potential at the center of the sphere?

An electron in a cathode ray tube is accelerated through a potential difference of 10 kV.What kinetic energy does the electron gain in the process? (e = 1.6 × 10−¹⁹ C)

A proton (+1.6 × 10−¹⁹ C) moves 10 cm on a path in the direction opposite to a uniform electric field of strength 3.0 N/C.How much work is done on the proton by the electrical field?

A 9.0-V battery moves 200 mC of charge through a circuit running from its positive terminal to its negative terminal.How much energy was delivered to the circuit?

At what distance from a point charge of 14 μC would the electrical potential be 4.2 × 10⁴ V? (k_e = 8.99 × 10⁹ N⋅m²/C²)

A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of −9.0 μC is at x = 1.00 m.What is the electric potential at the x = 0.50 m point? (k_e = 8.99 × 10⁹ N⋅m²/C²)

An electron (charge −1.6 × 10−¹⁹ C) moves on a path perpendicular to the direction of a uniform electric field of strength 3.0 N/C.How much work is done on the electron as it moves 20 cm?

If an electron is accelerated from rest through a potential difference of 300 V,find its approximate speed at the end of this process.(e = 1.6 × 10−¹⁹ C;m_e = 9.1 × 10−³¹ kg)

A uniform electric field,with a magnitude of 600 N/C,is directed parallel to the positive x axis.If the potential at x = 4.0 m is 1000 V,what is the potential at x = 2.0 m?

Two protons,each of charge 1.60 × 10−¹⁹ C,are 3.00 × 10−⁵ m apart.What is the change in potential energy if they are brought 1.00 × 10−⁵ m closer together? (k_e = 8.99 × 10⁹ N·m²/C²)

A uniform electric field,with a magnitude of 600 N/C,is directed parallel to the positive x axis.If the potential at x = 4.0 m is 1000 V,what is the change in potential energy of a proton as it moves from x = 4.0 m to x = 1.0 m? (q_p = 1.6 × 10−¹⁹ C)

Quiz 18: Direct-Current Circuits Sources of EMF: Part A

A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of −9.0 μC is at x = 1.0 m.At what point on the x axis is the electrical potential zero?

A proton (+1.6 × 10−19 C) moves 10 cm along the direction of an electric field of strength 4.0 N/C.The electrical potential difference between the proton's initial and ending points is:

An electron is released from rest at the negative plate of a parallel-plate capacitor.If the distance across the plates is 5.0 mm and the potential difference across the plates is 20 V,with what velocity does the electron hit the positive plate? (me = 9.1 × 10−31 kg,e = 1.6 × 10−19 C)

Two charges,each of charge q,are separated by a distance d.The distance between the charges is doubled.If the original potential energy of the arrangement was 40 mJ,by how much does the potential energy change when the charges are moved to the increased distance?

Find the electrical potential at 0.055 m from a point charge of 6.0 μC.(ke = 8.99 × 109 N⋅m2/C2)

An ion is released from rest and moves due to the force from an electric field from a position in the field having a potential of 14 V to a position having a potential of 8 V.The ion:

A solid conducting sphere of 10 cm radius has a net charge of 40 nC.If the potential at infinity is taken as zero,what is the potential at the center of the sphere?

An electron in a cathode ray tube is accelerated through a potential difference of 10 kV.What kinetic energy does the electron gain in the process? (e = 1.6 × 10−19 C)

A proton (+1.6 × 10−19 C) moves 10 cm on a path in the direction opposite to a uniform electric field of strength 3.0 N/C.How much work is done on the proton by the electrical field?

A 9.0-V battery moves 200 mC of charge through a circuit running from its positive terminal to its negative terminal.How much energy was delivered to the circuit?

At what distance from a point charge of 14 μC would the electrical potential be 4.2 × 104 V? (ke = 8.99 × 109 N⋅m2/C2)

A point charge of +3.0 μC is located at the origin of a coordinate system and a second point charge of −9.0 μC is at x = 1.00 m.What is the electric potential at the x = 0.50 m point? (ke = 8.99 × 109 N⋅m2/C2)

An electron (charge −1.6 × 10−19 C) moves on a path perpendicular to the direction of a uniform electric field of strength 3.0 N/C.How much work is done on the electron as it moves 20 cm?

If an electron is accelerated from rest through a potential difference of 300 V,find its approximate speed at the end of this process.(e = 1.6 × 10−19 C;me = 9.1 × 10−31 kg)