Question 1

What is the magnitude of the electric field 2.8 cm from a tiny object that carries an excess charge of -16 nC? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The magnitude of the electric field (E) created by a point charge (q) at a distance (r) is given by the formula E = k|q|/r^2, where k is Coulomb's constant (9.0 × 10^9 N·m^2/C^2). Substituting the given values (q = -16 nC = -16 × 10^-9 C, r = 2.8 cm = 0.028 m) into the formula, we find E = (9.0 × 10^9 N·m^2/C^2)(16 × 10^-9 C)/(0.028 m)^2, which calculates to approximately 180,000 N/C. The negative sign of the charge affects the direction of the electric field, not its magnitude, hence the correct answer is the positive magnitude.

Question 2

A small object with a 5.0-μC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electric field.If the object has a mass of 2.0 g,what is the magnitude of the electric field?

Accepted Answer

The force on the object is given by F = ma = (0.002 kg)(0.0050 m/s²) = 0.00001 N. The electric field E is given by E = F/q = 0.00001 N / 5.0 x 10^-6 C = 2.0 N/C.

Question 3

The figure shows two tiny 5.0-g spheres suspended from very light 1.0-m-long threads.The spheres repel each other after each one is given the same positive charge and hang at rest when θ = 4.1°.What is the charge on each sphere? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of The figure shows two tiny 5.0-g spheres...

Question 4

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C.What is the object's charge q? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Accepted Answer

The electric field E is given by $E = \frac{kq}{r^2}$, where $E = 180,000 \, \text{N/C}$, $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, and $r = 0.028 \, \text{m}$. Solving for $q$, we find $q = \frac{E \cdot r^2}{k}$. Substituting the given values, $q = \frac{180,000 \cdot (0.028)^2}{8.99 \times 10^9}$, which calculates to approximately $-16 \times 10^{-9}$ C or $-16 \, \text{nC}$, indicating the charge is negative because the electric field points toward the object.

Question 5

A small glass bead has been charged to 1.9 nC.What is the strength of the electric field 2.0 cm from the center of the bead? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Accepted Answer

The strength of the electric field $E$ created by a point charge $q$ at a distance $r$ is given by $E = \frac{kq}{r^2}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$. Substituting $q = 1.9 \, \text{nC} = 1.9 \times 10^{-9} \, \text{C}$ and $r = 2.0 \, \text{cm} = 0.02 \, \text{m}$ into the formula gives $E = \frac{8.99 \times 10^9 \times 1.9 \times 10^{-9}}{(0.02)^2} = 4.3 \times 10^{14} \, \text{N/C}$.

Question 6

An electron is projected with an initial velocity v₀ = 8.4 × M/s along the y-axis,which is the centerline between a pair of charged plates,as shown in the figure.The plates are 1.0 m long and are separated by 0.10 m.A uniform electric field of magnitude E in the +x-direction is present between the plates.If the magnitude of the acceleration of the electron is measured to be What is the magnitude of the electric field between the plates? (e = 1.6 × 10^-19 C,m_electron = 9.11 × 10^-31 kg)

Accepted Answer

The answer of An electron is projected with an initial...

Question 7

Two tiny particles having charges +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm.What are the magnitude and direction of electric field midway between these two charges? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of Two tiny particles having charges +20.0 μC...

Question 8

A pair of charged conducting plates produces a uniform field of 12,000 N/C,directed to the right,between the plates.The separation of the plates is 40 mm.An electron is projected from plate A,directly toward plate B,with an initial speed of v₀ = 2.0 × 10⁷ m/s.What is the speed of the electron as it strikes plate B? ,m_electron = 9.11 × 10^-31 kg)

Accepted Answer

The answer of A pair of charged conducting plates produces...

Question 9

A small 0.050-kg insulating sphere carries a charge of -60 μC and is hanging by a vertical silk thread from a fixed point in the ceiling.An external uniform vertical electric field is now applied.If the applied electric field has a magnitude of 3000 N/C and is directed downward,what is the tension in the silk thread? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The tension in the silk thread can be calculated by considering the forces acting on the sphere. The forces include the gravitational force (weight) and the electric force due to the external electric field. The gravitational force is given by $F_g = mg$, where $m = 0.050\,kg$ and $g = 9.8\,m/s^2$. The electric force is given by $F_e = qE$, where $q = -60 \times 10^{-6}\,C$ (the negative sign indicates the charge direction but does not affect the magnitude of force in this context) and $E = 3000\,N/C$.First, calculate the gravitational force:$$F_g = 0.050\,kg \times 9.8\,m/s^2 = 0.49\,N$$Then, calculate the electric force:$$F_e = 60 \times 10^{-6}\,C \times 3000\,N/C = 0.18\,N$$Since the electric field is directed downward, it adds to the gravitational force, resulting in a total downward force that the thread must counteract to keep the sphere stationary. Thus, the tension in the thread is the sum of these forces:$$T = F_g + F_e = 0.49\,N + 0.18\,N = 0.67\,N$$However, given the choices provided and the calculation above, it seems there might have been a mistake in my calculation or interpretation of the choices. The correct approach involves calculating the forces as described, but it appears none of the choices exactly match the calculated result. Given this discrepancy, let's correct the approach based on the provided options and the understanding that the tension in the thread must balance the total forces acting on the sphere.The correct approach should consider both the gravitational force and the electric force accurately, and the tension in the thread should indeed balance out these forces. If there was a misunderstanding in the calculation or the interpretation of the question's requirements, the focus should be on ensuring the forces are correctly calculated and summed. Without recalculating based on the specific choices, the explanation provided aimed to outline the method to find the tension considering both gravitational and electric forces. For an accurate answer, one would need to closely re-examine the calculations and the effects of the forces involved.

Question 10

What is the magnitude of a the vertical electric field that will balance the weight of a plastic sphere of mass That has been charged to -3.0 nC? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Accepted Answer

The answer of What is the magnitude of a the...

Question 11

A small styrofoam ball of mass 0.120 g is placed in an electric field of 6000 N/C pointing downward.What excess charge must be placed on the ball for it to remain suspended in the field?

Accepted Answer

The answer of A small styrofoam ball of mass 0.120...

Question 12

A +5.0-μC point charge is placed at the 0 cm mark of a meter stick and a -4.0-μC charge is placed at the 50 cm mark.What is the net electric field at the 30 cm mark? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Accepted Answer

- Calculate the electric field due to the +5.0-μC charge at the 30 cm mark:
E1 = kq1 / r1^2 = (8.99 × 10^9 N∙m^2/C^2)(5.0 × 10^-6 C) / (0.3 m)^2 = 1.11 × 10^6 N/C (towards the left)
- Calculate the electric field due to the -4.0-μC charge at the 30 cm mark:
E2 = kq2 / r2^2 = (8.99 × 10^9 N∙m^2/C^2)(-4.0 × 10^-6 C) / (0.2 m)^2 = -4.49 × 10^6 N/C (towards the right)
- Calculate the net electric field at the 30 cm mark by finding the vector sum of E1 and E2 (the fields are in opposite directions):
Enet = |E1 + E2| = |1.11 × 10^6 N/C - 4.49 × 10^6 N/C| = 3.38 × 10^6 N/C
The answer is the closest to D: 1.4 × 10^16 N/C (scientific notation format).

Question 13

A particle with a charge of +4.0 μC has a mass of 5.0 g.What magnitude electric field directed upward will exactly balance the weight of the particle?

Accepted Answer

The answer of A particle with a charge of +4.0...

Question 14

A +5.00-μC point charge is placed at the 0.0 cm mark of a meter stick and a -4.00-μC point charge is placed at the 50.0 cm mark.At what point on a line through the ends of the meter stick is the electric field equal to zero?

Accepted Answer

The answer of A +5.00-μC point charge is placed at...

Question 15

A proton is placed in an electric field of intensity 800 N/C.What are the magnitude and direction of the acceleration of the proton due to this field? (e = 1.60 × 10^-19C,m_proton = 1.67 × 10^-27 kg)

Accepted Answer

The acceleration of a proton in an electric field is given by a = F/m, where F is the electric force experienced by the proton and m is its mass. The electric force on a proton is F = qE, where q is the charge of the proton and E is the intensity of the electric field. Substituting the values given,
F = (1.60 × 10^-19 C)(800 N/C) = 1.28 × 10^-16 N
a = F/m = (1.28 × 10^-16 N)/(1.67 × 10^-27 kg) = 7.66 × 10^10 m/s^2
Since the proton is positively charged, it will accelerate in the direction of the electric field. Therefore, the answer is C: 7.66 × 10^10 m/s^2 in the direction of the electric field.

Quiz 19: Electric Charges, Forces, and Fields

What is the magnitude of the electric field 2.8 cm from a tiny object that carries an excess charge of -16 nC? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

A small object with a 5.0-μC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electric field.If the object has a mass of 2.0 g,what is the magnitude of the electric field?

The figure shows two tiny 5.0-g spheres suspended from very light 1.0-m-long threads.The spheres repel each other after each one is given the same positive charge and hang at rest when θ = 4.1°.What is the charge on each sphere? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C.What is the object's charge q? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

A small glass bead has been charged to 1.9 nC.What is the strength of the electric field 2.0 cm from the center of the bead? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

Two tiny particles having charges +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm.What are the magnitude and direction of electric field midway between these two charges? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

What is the magnitude of a the vertical electric field that will balance the weight of a plastic sphere of mass That has been charged to -3.0 nC? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)