Question 1

Two charges Q₁ and Q₂ are a distance d apart. If the electric field is zero at a distance of 3d/4 from Q₁ towards Q₂), then what is the relation between Q₁ and Q₂?

Accepted Answer

The electric field is zero at a point between two charges when the magnitudes of the electric fields due to each charge are equal at that point. The electric field due to a charge is given by $E = k\frac{Q}{r^2}$, where $E$ is the electric field, $k$ is Coulomb's constant, $Q$ is the charge, and $r$ is the distance from the charge. If the electric field is zero at a distance of $3d/4$ from $Q_{₁}$ towards $Q_{₂}$, then the distance from $Q_{₂}$ is $d - 3d/4 = d/4$. Setting the magnitudes of the electric fields equal to each other and solving for the ratio of the charges gives $Q_{₁} = 9Q_{₂}$, because the distance from $Q_{₁}$ is squared in the denominator, making the ratio of the distances squared (i.e., $(3d/4)^2 / (d/4)^2 = 9$).

Question 2

Three charges, each of Q = 3.2 × 10^-¹⁹ C, are arranged at three of the corners of a 20-nm square as shown. The magnitude of the electric field at D, the fourth corner of the square, is approximately

Accepted Answer

The answer of  Three charges, each of Q =...

Question 3

An electron is moving horizontally east in an electric field that points vertically upward. The electric force on the proton is

Accepted Answer

An electron is negatively charged, so it experiences a force opposite in direction to the electric field lines. In this case, since the electric field is pointing upward, the force on the electron is downward. Therefore, the answer is (C) downward.

Question 4

A conducting sphere has a net charge of -q and of mass m is suspended from the ceiling by a light string. A uniform electric field E is applied vertical down on the sphere.If m = 1 g, q = 1 μC, and E = 5000 N/C, the tension T in the string is

Accepted Answer

The net force on the sphere is the gravitational force minus the electric force:
Fnet = mg - qE
Plugging in the values given, we get:
Fnet = (0.001 kg)(9.81 m/s^2) - (1 × 10^-6 C)(5000 N/C) = 4.81 × 10^-3 N
Since the sphere is suspended by a string, the tension in the string must balance the net force:
T = Fnet = 4.81 × 10^-3 N

Question 5

Two charges Q1 and Q2 are 8.0 cm apart. Charge Q1 = 5.0 nC and Q2 = -5.0 nC. The magnitude of the electric field at point P, 3.0 cm from the midpoint of the line joining Q1 and Q2, is

Accepted Answer

The answer of  Two charges Q1 and Q2 are...

Question 6

Charges Q₁ = -q and Q₂ = +4q are placed as shown. Of the five positions indicated by the numbered dots, the one at which the electric field E is zero is

Accepted Answer

The answer of Charges Q₁ = -q and Q₂...

Question 7

Two charges of the same magnitude and sign are placed a certain distance apart. There is only one point in space near them where the electric field is zero. Which, if any, of the following statements about that point is true?

Accepted Answer

By symmetry, the point where the electric field is zero must be on the line joining the charges. Since the charges have the same sign, the electric field points away from both of them. Therefore, the only way for the electric fields to cancel each other out is if the point is on the line joining the charges and between the charges.

Question 8

If all the charges are 15 cm from the origin the crossing point of the vertical and horizontal lines) in the above figure and Q = +3.0 μC, then calculate the magnitude of the net force on a charge of +Q placed at the origin.

Accepted Answer

The answer of  If all the charges are 15...

Question 9

A conducting sphere has a net charge of -q and of mass m is suspended from the ceiling by a light string. A uniform electric field E is applied vertical down on the sphere.The tension T in the string is ____ the weight mg.

Accepted Answer

The sphere will experience an upward force due to the tension in the string that is equal and opposite to the force of gravity (weight). However, the electric field E will also exert a downward force on the sphere due to its negative charge. Since the string is light and not likely to have significant stretch or bending, we can assume that the tension in the string is primarily due to the weight of the sphere. Thus, if the electric field E is strong enough to overcome the weight of the sphere, the tension in the string will be less than the weight (and vice versa). Therefore, the tension T in the string is expected to be less than the weight mg in this scenario.

Question 10

A particular nucleus of the element plutonium contains 94 protons and 150 neutrons. What is the magnitude of the electric field at a distance of 0.563 × 10^-¹² m from the nucleus?

Accepted Answer

The answer of A particular nucleus of the element plutonium...

Question 11

A proton is moving horizontally north in an electric field that points vertically upward. The electric force on the proton is

Accepted Answer

The electric force is in the direction of the electric field, which is vertically upward. Since the proton is moving horizontally to the north, the direction of the electric force will be perpendicular to the direction of motion, which is upward. Therefore, the electric force on the proton is upward.

Question 12

The electric field at point A is zero. What is charge Q₁?

Accepted Answer

According to the question, the electric field at point A is zero. This means that the electric field due to Q₁ and QTB7291_00 at point A must be equal and opposite in direction. Since the distance between Q₁ and A is smaller than that between QTB7291_00 and A, the magnitude of the charge at Q₁ must be larger than that at QTB7291_00. Thus, the only option where Q₁ has a larger charge magnitude than QTB7291_00 is option A (+32 μC). Therefore, the answer is (A).

Question 13

In the diagram, Q₁ = 6.0 μC and Q₂ = -6.0 μC. The electric field at point 2, 0) is

Accepted Answer

The answer of In the diagram, Q₁ = 6.0...

Question 14

The direction of the electric field at a point is the same as

Accepted Answer

The direction of the electric field at a point is defined as the direction of the force that would be exerted on a positive charge placed at that point. Since a proton is positively charged, the force on a proton placed at that point would be in the same direction as the electric field.

Question 15

Three charges Q1, Q2, and Q3, each equal to 6.4 × 10^-¹⁹ C, are in a straight line. The distance between neighboring charges is 60 nm. The magnitude of the electric field at P, which is 80 nm from Q2 on a line at right angles to the line between Q1 and Q3, is

Accepted Answer

The electric field at point P will be the vector sum of the electric fields at that point due to the individual charges Q1, Q2, and Q3. We can calculate the magnitude of the electric field due to each charge using the formula:

E = kQ/r^2

where k is Coulomb's constant (9 × 10^9 N·m^2/C^2), Q is the charge, and r is the distance from the charge to the point P.

For Q1 and Q3, the distance is:

sqrt((80 nm)^2 + (30 nm)^2) = 86.8 nm

Therefore, the magnitude of the electric field due to Q1 and Q3 at point P is:

E1 = E3 = kQ/r^2 = 9 × 10^9 × (6.4 × 10^-10 C)/(86.8 × 10^-9 m)^2 = 1.9 × 10^5 N/C

For Q2, the distance is simply:

60 nm

Therefore, the magnitude of the electric field due to Q2 at point P is:

E2 = kQ/r^2 = 9 × 10^9 × (6.4 × 10^-10 C)/(60 × 10^-9 m)^2 = 2.2 × 10^5 N/C

Note that the electric field due to Q2 is larger than that due to Q1 or Q3 because point P is closer to Q2.

Now we need to add the electric fields due to each charge vectorially. Since the charges are in a straight line and Q1 and Q3 are equidistant from Q2, the electric fields due to Q1 and Q3 will cancel out in the x-direction (i.e. along the line connecting the charges), leaving only the y-component of those fields.

We can use trigonometry to find the y-component of each electric field:

sin θ = (30 nm)/86.8 nm = 0.345

cos θ = (80 nm)/86.8 nm = 0.921

The y-component of E1 and E3 is:

E1y = E1 sin θ = 1.9 × 10^5 × 0.345 = 6.5 × 10^4 N/C

E3y = E3 sin θ = 1.9 × 10^5 × 0.345 = 6.5 × 10^4 N/C

The x-component of E2 and the y-component of all three electric fields are in the same direction, so we can simply add them up:

Ey = E1y + E2 + E3y = (6.5 × 10^4 + 2.2 × 10^5 + 6.5 × 10^4) N/C = 3.4 × 10^5 N/C

Finally, we can use the Pythagorean theorem to find the magnitude of the total electric field at point P:

E = sqrt((Ex)^2 + (Ey)^2) = sqrt((0)^2 + (3.4 × 10^5 N/C)^2) = 3.4 × 10^5 N/C

So the answer is D, 1.9 × 10^5 N/C.

Question 16

Two charges Q₁ and Q₂ are distance d apart. If the electric field is zero at distance 3d/2 from Q₁ and d/2 from Q₂, along the line joining Q₁ and Q₂, then what is the relation between Q₁ and Q₂?

Accepted Answer

The answer of Two charges Q₁ and Q₂ are distance...

Question 17

Two point charges of unknown magnitude and sign are a distance d apart. If the electric field strength is zero at a point between them on the line joining them, you can conclude that

Accepted Answer

If the electric field strength is zero at a point between the two charges, it means the field due to one charge is equal and opposite to the field due to the other charge at that point. This can only happen if the charges are equal in magnitude but opposite in sign, as only then the fields due to the charges cancel each other.Here, there are no other possibilities for charges so the answer has to be A.

Question 18

An electric field with a magnitude of 6.0 × 10⁴ N/C is directed parallel to the positive y-axis. A particle with a charge q = 4.8 × 10^-¹⁹ C is moving along the x-axis with a speed v = 3.0 × 10⁶ m/s. The force on the charge is approximately

Accepted Answer

The force on a charged particle moving in an electric field is given by $F=qE$, where $q$ is the charge of the particle and $E$ is the electric field. In this case, the force is $F=(4.8	imes10^{-19}	ext{ C})(6.0	imes10^{14}	ext{ N/C})=2.9	imes10^{-5}	ext{ N}$. Since the electric field is parallel to the y-axis, and the particle is moving along the x-axis, the force will be perpendicular to both, i.e. in the y-direction. Thus, the answer is B, 2.9 × 10$^{-5}$ N in the y direction.

Question 19

Three positive and equal charges Q1, Q2, and Q3 are at the corners of an equilateral triangle as shown. Point P is at the midpoint of the line between Q1 and Q3. The electric field at P is

Accepted Answer

The answer of  Three positive and equal charges Q1,...

Question 20

Three charges are located at 100-m intervals along a horizontal line: a charge of -3.0 C on the left, 2.0 C in the middle, and 1.0 C on the right. What is the electric field E on the horizontal line halfway between the -3.0 C and 2.0 C charges?

Accepted Answer

To find the electric field at the midpoint between the -3.0 C and 2.0 C charges, we can use the principle of superposition to find the electric fields due to each individual charge and then add them together. The electric field due to each charged particle is given by Coulomb's law:
E = k * q / r^2
where k is Coulomb's constant, q is the charge, and r is the distance from the charge.

The electric field due to the -3.0 C charge at the midpoint is:
E1 = k * (-3.0 C) / (50 m)^2 = -1.08 × 10^8 N/C to the right

The electric field due to the 2.0 C charge at the midpoint is:
E2 = k * (2.0 C) / (50 m)^2 = 3.60 × 10^7 N/C to the left

The net electric field at the midpoint is the vector sum of these two fields:
E = E1 + E2 = (-1.08 + 0.36) × 10^8 N/C = -0.72 × 10^8 N/C = -1.8 × 10^7 N/C to the left.

Therefore, the answer is C, 1.8 × 10^17 N/C to the left.

Quiz 16: Electrostatics I: Electric Charge, Forces, and Fields

Two charges Q₁ and Q₂ are a distance d apart. If the electric field is zero at a distance of 3d/4 from Q₁ towards Q₂) , then what is the relation between Q₁ and Q₂?

Three charges, each of Q = 3.2 × 10^-¹⁹ C, are arranged at three of the corners of a 20-nm square as shown. The magnitude of the electric field at D, the fourth corner of the square, is approximately

An electron is moving horizontally east in an electric field that points vertically upward. The electric force on the proton is

A conducting sphere has a net charge of -q and of mass m is suspended from the ceiling by a light string. A uniform electric field E is applied vertical down on the sphere.If m = 1 g, q = 1 μC, and E = 5000 N/C, the tension T in the string is

Two charges Q1 and Q2 are 8.0 cm apart. Charge Q1 = 5.0 nC and Q2 = -5.0 nC. The magnitude of the electric field at point P, 3.0 cm from the midpoint of the line joining Q1 and Q2, is

Charges Q₁ = -q and Q₂ = +4q are placed as shown. Of the five positions indicated by the numbered dots, the one at which the electric field E is zero is

Two charges of the same magnitude and sign are placed a certain distance apart. There is only one point in space near them where the electric field is zero. Which, if any, of the following statements about that point is true?

If all the charges are 15 cm from the origin the crossing point of the vertical and horizontal lines) in the above figure and Q = +3.0 μC, then calculate the magnitude of the net force on a charge of +Q placed at the origin.

A conducting sphere has a net charge of -q and of mass m is suspended from the ceiling by a light string. A uniform electric field E is applied vertical down on the sphere.The tension T in the string is ____ the weight mg.

A particular nucleus of the element plutonium contains 94 protons and 150 neutrons. What is the magnitude of the electric field at a distance of 0.563 × 10^-¹² m from the nucleus?

A proton is moving horizontally north in an electric field that points vertically upward. The electric force on the proton is

The electric field at point A is zero. What is charge Q₁?

In the diagram, Q₁ = 6.0 μC and Q₂ = -6.0 μC. The electric field at point 2, 0) is

The direction of the electric field at a point is the same as

Three charges Q1, Q2, and Q3, each equal to 6.4 × 10^-¹⁹ C, are in a straight line. The distance between neighboring charges is 60 nm. The magnitude of the electric field at P, which is 80 nm from Q2 on a line at right angles to the line between Q1 and Q3, is

Two charges Q₁ and Q₂ are distance d apart. If the electric field is zero at distance 3d/2 from Q₁ and d/2 from Q₂, along the line joining Q₁ and Q₂, then what is the relation between Q₁ and Q₂?

Two point charges of unknown magnitude and sign are a distance d apart. If the electric field strength is zero at a point between them on the line joining them, you can conclude that

An electric field with a magnitude of 6.0 × 10⁴ N/C is directed parallel to the positive y-axis. A particle with a charge q = 4.8 × 10^-¹⁹ C is moving along the x-axis with a speed v = 3.0 × 10⁶ m/s. The force on the charge is approximately

Three positive and equal charges Q1, Q2, and Q3 are at the corners of an equilateral triangle as shown. Point P is at the midpoint of the line between Q1 and Q3. The electric field at P is

Three charges are located at 100-m intervals along a horizontal line: a charge of -3.0 C on the left, 2.0 C in the middle, and 1.0 C on the right. What is the electric field E on the horizontal line halfway between the -3.0 C and 2.0 C charges?

Quiz 16: Electrostatics I: Electric Charge, Forces, and Fields

Two charges Q1 and Q2 are a distance d apart. If the electric field is zero at a distance of 3d/4 from Q1 towards Q2) , then what is the relation between Q1 and Q2?

Three charges, each of Q = 3.2 × 10-19 C, are arranged at three of the corners of a 20-nm square as shown. The magnitude of the electric field at D, the fourth corner of the square, is approximately

An electron is moving horizontally east in an electric field that points vertically upward. The electric force on the proton is

A conducting sphere has a net charge of -q and of mass m is suspended from the ceiling by a light string. A uniform electric field E is applied vertical down on the sphere.If m = 1 g, q = 1 μC, and E = 5000 N/C, the tension T in the string is

Two charges Q1 and Q2 are 8.0 cm apart. Charge Q1 = 5.0 nC and Q2 = -5.0 nC. The magnitude of the electric field at point P, 3.0 cm from the midpoint of the line joining Q1 and Q2, is

Charges Q1 = -q and Q2 = +4q are placed as shown. Of the five positions indicated by the numbered dots, the one at which the electric field E is zero is

Two charges of the same magnitude and sign are placed a certain distance apart. There is only one point in space near them where the electric field is zero. Which, if any, of the following statements about that point is true?

If all the charges are 15 cm from the origin the crossing point of the vertical and horizontal lines) in the above figure and Q = +3.0 μC, then calculate the magnitude of the net force on a charge of +Q placed at the origin.

A conducting sphere has a net charge of -q and of mass m is suspended from the ceiling by a light string. A uniform electric field E is applied vertical down on the sphere.The tension T in the string is ____ the weight mg.

A particular nucleus of the element plutonium contains 94 protons and 150 neutrons. What is the magnitude of the electric field at a distance of 0.563 × 10-12 m from the nucleus?

A proton is moving horizontally north in an electric field that points vertically upward. The electric force on the proton is

The electric field at point A is zero. What is charge Q1?

In the diagram, Q1 = 6.0 μC and Q2 = -6.0 μC. The electric field at point 2, 0) is

The direction of the electric field at a point is the same as

Three charges Q1, Q2, and Q3, each equal to 6.4 × 10-19 C, are in a straight line. The distance between neighboring charges is 60 nm. The magnitude of the electric field at P, which is 80 nm from Q2 on a line at right angles to the line between Q1 and Q3, is

Two charges Q1 and Q2 are distance d apart. If the electric field is zero at distance 3d/2 from Q1 and d/2 from Q2, along the line joining Q1 and Q2, then what is the relation between Q1 and Q2?

Two point charges of unknown magnitude and sign are a distance d apart. If the electric field strength is zero at a point between them on the line joining them, you can conclude that

An electric field with a magnitude of 6.0 × 104 N/C is directed parallel to the positive y-axis. A particle with a charge q = 4.8 × 10-19 C is moving along the x-axis with a speed v = 3.0 × 106 m/s. The force on the charge is approximately

Three positive and equal charges Q1, Q2, and Q3 are at the corners of an equilateral triangle as shown. Point P is at the midpoint of the line between Q1 and Q3. The electric field at P is

Three charges are located at 100-m intervals along a horizontal line: a charge of -3.0 C on the left, 2.0 C in the middle, and 1.0 C on the right. What is the electric field E on the horizontal line halfway between the -3.0 C and 2.0 C charges?

Two charges Q₁ and Q₂ are a distance d apart. If the electric field is zero at a distance of 3d/4 from Q₁ towards Q₂) , then what is the relation between Q₁ and Q₂?

Three charges, each of Q = 3.2 × 10^-¹⁹ C, are arranged at three of the corners of a 20-nm square as shown. The magnitude of the electric field at D, the fourth corner of the square, is approximately

Charges Q₁ = -q and Q₂ = +4q are placed as shown. Of the five positions indicated by the numbered dots, the one at which the electric field E is zero is

A particular nucleus of the element plutonium contains 94 protons and 150 neutrons. What is the magnitude of the electric field at a distance of 0.563 × 10^-¹² m from the nucleus?

The electric field at point A is zero. What is charge Q₁?

In the diagram, Q₁ = 6.0 μC and Q₂ = -6.0 μC. The electric field at point 2, 0) is

Three charges Q1, Q2, and Q3, each equal to 6.4 × 10^-¹⁹ C, are in a straight line. The distance between neighboring charges is 60 nm. The magnitude of the electric field at P, which is 80 nm from Q2 on a line at right angles to the line between Q1 and Q3, is

Two charges Q₁ and Q₂ are distance d apart. If the electric field is zero at distance 3d/2 from Q₁ and d/2 from Q₂, along the line joining Q₁ and Q₂, then what is the relation between Q₁ and Q₂?

An electric field with a magnitude of 6.0 × 10⁴ N/C is directed parallel to the positive y-axis. A particle with a charge q = 4.8 × 10^-¹⁹ C is moving along the x-axis with a speed v = 3.0 × 10⁶ m/s. The force on the charge is approximately