Question 1

If Q = 25 μC,q = 10 μC,and L = 40 cm in the figure,what is the magnitude of the electrostatic force on q? 
A)28 N
B)22 N
C)20 N
D)14 N
E)10 N

Accepted Answer

C

Question 2

If a = 3.0 mm,b = 4.0 mm,Q₁ = 60 nC,Q₂ = 80 nC,and q = 24 nC in the figure,what is the magnitude of the electric force on q? A)2.7 N B)1.9 N C)2.3 N D)1.5 N E)0.52 N

Accepted Answer

D

Question 3

A point charge Q is placed on the x axis at x = −2.0 m.A second point charge,−Q,is placed at x = 1.0 m.If Q = 60 μC,what is the magnitude of the electrostatic force on a 40-μC charge placed at the origin?
A)16 N
B)27 N
C)32 N
D)11 N
E)3.0 N

Accepted Answer

B

Question 4

A particle (m = 50 g,q = 5.0 μC)is released from rest when it is 50 cm from a second particle (Q = −20 μC).Determine the magnitude of the initial acceleration of the 50-g particle. A)54 m/s² B)90 m/s² C)72 m/s² D)65 m/s² E)36 m/s²

Accepted Answer

The initial acceleration of the 50-g particle can be determined using Coulomb's law and Newton's law of motion. The force acting on the 50-g particle due to the second particle is:

$$F = \frac{kq_1q_2}{r^2}$$

where k is Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them. Substituting the given values:

$$F = \frac{(9	imes10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2)(5.0	imes10^{-6}	ext{ C})(-20	imes10^{-6}	ext{ C})}{(0.50	ext{ m})^2} = -720 	ext{ N}$$

Since the force is negative, it implies that the particles are attracted towards each other. By Newton's second law of motion, the acceleration of the 50-g particle is:

$$a = \frac{F}{m} = \frac{-720	ext{ N}}{0.050	ext{ kg}} = -14400 	ext{ m/s}^2$$

Taking the magnitude of the acceleration:

$$|a| = |-14400	ext{ m/s}^2| = 14400	ext{ m/s}^2$$

Therefore, the magnitude of the initial acceleration of the 50-g particle is 14400 m/s^2, which is closest to 72 m/s^2 (choice C).

Question 5

A point charge (−5.0 μC)is placed on the x axis at x = 4.0 cm,and a second charge (+5.0 μC)is placed on the x axis at x = −4.0 cm.What is the magnitude of the electric force on a third charge (+2.5 μC)placed on the y axis at y = 3.0 cm?
A)90 N
B)45 N
C)54 N
D)72 N
E)36 N

Accepted Answer

The answer of A point charge (−5.0 μC)is placed on...

Question 6

Identical point charges Q are placed at each of the four corners of a 3.0 m × 4.0 m rectangle.If Q = 40 μC,what is the magnitude of the electrostatic force on any one of the charges?
A)3.0 N
B)2.4 N
C)1.8 N
D)3.7 N
E)2.0 N

Accepted Answer

The magnitude of the electrostatic force between two point charges Q separated by a distance r is given by Coulomb's law: F = (1/4πε₀)(Q₁Q₂/r²), where ε₀ is the permittivity of free space. In this case, each of the charges Q experiences the same force due to symmetry. The distance between adjacent charges is either 3 m or 4 m, depending on the direction of the force. Using the distance formula for the diagonal of the rectangle, we find that the distance between opposite charges is 5 m. Substituting Q = 40 μC, r = 5 m, and ε₀ = 8.85 x 10⁻¹² C²/Nm², we get F = (1/4πε₀)(40x10⁻⁶)²/(5)² ≈ 2.4 N. Therefore, the answer is B.

Question 7

Three point charges,two positive and one negative,each having a magnitude of 20 μC are placed at the vertices of an equilateral triangle (30 cm on a side).What is the magnitude of the electrostatic force on the negative charge?
A)80 N
B)40 N
C)69 N
D)57 N
E)75 N

Accepted Answer

We can solve this problem using Coulomb's law, which states that the electrostatic force between two point charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Using this law, we can find the force on the negative charge due to each of the other charges separately, and then add them up vectorially. 
Let's choose the negative charge at the origin, and the positive charges at (15 cm, 15√3 cm) and (-15 cm, 15√3 cm) respectively. The distance between the charges is 30 cm, so the force between each pair of charges is given by:
F = kq1q2/r^2
where k is the Coulomb constant (9×10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them. For the positive charges, q1=q2=20 μC, and r=30 cm.
Fp = kq1q2/r^2 = (9×10^9)(20×10^-6)^2/(0.3)^2 = 600 N
Since the positive charges are identical, the vector sum of the forces on the negative charge due to them is just the sum of their individual forces, directed along the line connecting them:
Fp,total = 2Fp sin(60°) = 2×600×(√3/2) = 1044 N
For the negative charge, q=-20 μC, and the distance to each of the positive charges is also 30 cm. However, the direction of the force is along the line connecting the negative charge to each positive charge, but in the opposite direction. Therefore, the vector sum of the forces on the negative charge due to the positive charges is:
Fp,total = -2Fp sin(60°) = -2×600×(√3/2) = -1044 N
The total force on the negative charge is then the vector sum of the forces due to the positive charges and the negative charge:
Ftotal = Fp,total + Fn,total
where Fn,total is the force on the negative charge due to the other negative charge. Since the charges are identical in magnitude, but opposite in sign, the force between them is attractive and is given by:
Fn = kq1q2/r^2 = (9×10^9)(20×10^-6)^2/(0.3)^2 = 600 N
The direction of this force is along the line connecting the two negative charges. Since this line is perpendicular to the line connecting the negative charge to either positive charge, the vector sum of the forces is just the magnitude of the force due to the other negative charge:
Fn,total = Fn = 600 N
Finally, the total force on the negative charge is:
Ftotal = -1044 + 600 = -444 N
The magnitude of this force is just the absolute value, which is 444 N. Therefore, the answer is (C) 69 N.

Question 8

In the figure,if Q = 30 μC,q = 5.0 μC,and d = 30 cm,what is the magnitude of the electrostatic force on q? 
A)15 N
B)23 N
C)zero
D)7.5 N
E)38 N

Accepted Answer

The magnitude of the electrostatic force acting on q is given by:

F = (1/4πε0)(Qq/d^2), where ε0 is the permittivity of free space.

Plugging in the given values, we get:

F = (1/4πε0)(30×10^-6 C × 5.0×10^-6 C)/(0.3 m)^2

F ≈ 7.5 N

Therefore, the correct answer is choice D, 7.5 N.

Question 9

A point charge Q is placed at the origin.A second charge,2Q,is placed on the x axis at x = −3.0 m.If Q = 50 μC,what is the magnitude of the electrostatic force on a third point charge,−Q,placed on the y axis at y = +4.0 m?
A)2.5 N
B)3.0 N
C)3.7 N
D)4.4 N
E)1.8 N

Accepted Answer

The answer of A point charge Q is placed at...

Question 10

If a = 3.0 mm,b = 4.0 mm,Q₁ = 60 nC,Q₂ = −80 nC,and q = 36 nC in the figure,what is the magnitude of the electric force on q? A)5.0 N B)4.4 N C)3.8 N D)5.7 N E)0.60 N

Accepted Answer

The answer of If a = 3.0 mm,b = 4.0...

Question 11

A charge of +80 μC is placed on the x axis at x = 0.A second charge of −50 μC is placed on the x axis at x = 50 cm.What is the magnitude of the electrostatic force on a third charge of 4.0 μC placed on the x axis at x = 30 cm?
A)13 N
B)77 N
C)39 N
D)25 N
E)45 N

Accepted Answer

The answer of A charge of +80 μC is placed...

Question 12

A point charge Q is placed on the x axis at x = 2.0 m.A second point charge,−Q,is placed at x = 3.0 m.If Q = 40 μC,what is the magnitude of the electrostatic force on a 30-μC charge placed at the origin?
A)7.2 N
B)3.9 N
C)1.5 N
D)14 N
E)8.1 N

Accepted Answer

The answer of A point charge Q is placed on...

Question 13

Three identical point charges Q are placed at the vertices of an equilateral triangle (length of each side = 2.0 m).If Q = 60 μC,what is the magnitude of the electrostatic force on any one of the charges?
A)25 N
B)19 N
C)14 N
D)22 N
E)16 N

Accepted Answer

The answer of Three identical point charges Q are placed...

Question 14

A particle (charge = +40 μC)is located on the x axis at the point x = −20 cm,and a second particle (charge = −50 μC)is placed on the x axis at x = +30 cm.What is the magnitude of the total electrostatic force on a third particle (charge = −4.0 μC)placed at the origin (x = 0)?
A)41 N
B)16 N
C)56 N
D)35 N
E)72 N

Accepted Answer

The answer of A particle (charge = +40 μC)is located...

Question 15

Three point charges,two positive and one negative,each having a magnitude of 20 μC are placed at the vertices of an equilateral triangle (30 cm on a side).What is the magnitude of the electrostatic force on one of the positive charges?
A)69 N
B)40 N
C)80 N
D)57 N
E)20 N

Accepted Answer

The answer of Three point charges,two positive and one negative,each...

Question 16

Three point charges are positioned on the x axis.If the charges and corresponding positions are +32 μC at x = 0,+20 μC at x = 40 cm,and −60 μC at x = 60 cm,what is the magnitude of the electrostatic force on the +32-μC charge?
A)84 N
B)12 N
C)36 N
D)50 N
E)48 N

Accepted Answer

To calculate the electrostatic force on the +32 μC charge, we need to calculate the force due to the other two charges, using Coulomb's law:

$F_{1}=k\frac{q_{1}q_{2}}{r_{1}^{2}}$

$F_{2}=k\frac{q_{1}q_{3}}{r_{2}^{2}}$

where $k=9	imes10^9\frac{N\cdot m^2}{C^2}$ is the Coulomb constant, $q_{1}=+32\mu C$ is the charge at the origin, $q_{2}=+20\mu C$ is the charge at x=40 cm, $q_{3}=-60\mu C$ is the charge at x=60 cm, $r_1=40 cm=0.4 m$ is the distance between charges at the origin and x=40 cm, and $r_2=60 cm=0.6 m$ is the distance between charges at the origin and x=60 cm.

$F_{1}=9	imes10^9\frac{(32	imes10^{-6}C)(20	imes10^{-6}C)}{(0.4m)^{2}}=36N$

$F_{2}=9	imes10^9\frac{(32	imes10^{-6}C)(-60	imes10^{-6}C)}{(0.6m)^{2}}=-24N$

The total electrostatic force on the +32 μC charge is the vector sum of these two forces:

$F=F_{1}+F_{2}=36N-24N=12N$

So, the answer is B) 12 N.

Question 17

If a = 3.0 mm,b = 4.0 mm,Q₁ = 60 nC,Q₂ = 80 nC,and q = 32 nC in the figure,what is the magnitude of the electric force on q? A)1.6 N B)1.3 N C)1.9 N D)2.2 N E)0.040 N

Accepted Answer

Using Coulomb's law, we can calculate the electric force on q: F = k * (Q₁*q/a^2 + Q₂*q/b^2) where k is Coulomb's constant: k = 9 * 10^9 N*m^2/C^2 Plugging in the given values, we get: F = 9 * 10^9 * (60 * 32 / 3^2 + 80 * 32 / 4^2) F ≈ 1.3 N Therefore, the answer is B.

Question 18

If a = 3.0 mm,b = 4.0 mm,Q₁ = −60 nC,Q₂ = 80 nC,and q = 30 nC in the figure,what is the magnitude of the electric force on q? A)1.4 N B)1.0 N C)1.7 N D)2.0 N E)0.50 N

Accepted Answer

The answer of If a = 3.0 mm,b = 4.0...

Question 19

Each of two small non-conducting spheres is charged positively,the combined charge being 40 μC.When the two spheres are 50 cm apart,each sphere is repelled from the other by a force of magnitude 2.0 N.Determine the magnitude of the smaller of the two charges.
A)1.4 μC
B)1.1 μC
C)2.0 μC
D)3.3 μC
E)17 μC

Accepted Answer

The answer of Each of two small non-conducting spheres is...

Question 20

A point charge Q is placed on the x axis at the origin.An identical point charge is placed on the x axis at x = −1.0 m and another at x = +1.0 m.If Q = 40 μC,what is the magnitude of the electrostatic force on the charge at x = +1.0 m?
A)29 N
B)14 N
C)11 N
D)18 N
E)7.0 N

Accepted Answer

The electrostatic force between two point charges can be calculated using Coulomb's law: 
$F=k\frac{Q_1Q_2}{r^2}$ 
where k is Coulomb's constant, Q1 and Q2 are the charges, and r is the distance between them. 
In this case, we have three charges: Q1 at the origin, Q2 at -1.0 m, and Q3 at +1.0 m. The force on Q3 can be calculated by finding the electric field due to Q1 and Q2 at that point, and then multiplying by the charge Q3: 
$F=kQ_1Q_3\frac{1}{(1.0)^2}+kQ_2Q_3\frac{1}{(2.0)^2}=kQ_3(40\mu C)\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ 
$F\approx18 N$ 
Therefore, the answer is D.

Quiz 23: Electric Fields

If Q = 25 μC,q = 10 μC,and L = 40 cm in the figure,what is the magnitude of the electrostatic force on q?

If a = 3.0 mm,b = 4.0 mm,Q₁ = 60 nC,Q₂ = 80 nC,and q = 24 nC in the figure,what is the magnitude of the electric force on q?

A point charge Q is placed on the x axis at x = −2.0 m.A second point charge,−Q,is placed at x = 1.0 m.If Q = 60 μC,what is the magnitude of the electrostatic force on a 40-μC charge placed at the origin?

A particle (m = 50 g,q = 5.0 μC) is released from rest when it is 50 cm from a second particle (Q = −20 μC) .Determine the magnitude of the initial acceleration of the 50-g particle.

A point charge (−5.0 μC) is placed on the x axis at x = 4.0 cm,and a second charge (+5.0 μC) is placed on the x axis at x = −4.0 cm.What is the magnitude of the electric force on a third charge (+2.5 μC) placed on the y axis at y = 3.0 cm?

Identical point charges Q are placed at each of the four corners of a 3.0 m × 4.0 m rectangle.If Q = 40 μC,what is the magnitude of the electrostatic force on any one of the charges?

Three point charges,two positive and one negative,each having a magnitude of 20 μC are placed at the vertices of an equilateral triangle (30 cm on a side) .What is the magnitude of the electrostatic force on the negative charge?

In the figure,if Q = 30 μC,q = 5.0 μC,and d = 30 cm,what is the magnitude of the electrostatic force on q?

A point charge Q is placed at the origin.A second charge,2Q,is placed on the x axis at x = −3.0 m.If Q = 50 μC,what is the magnitude of the electrostatic force on a third point charge,−Q,placed on the y axis at y = +4.0 m?

If a = 3.0 mm,b = 4.0 mm,Q₁ = 60 nC,Q₂ = −80 nC,and q = 36 nC in the figure,what is the magnitude of the electric force on q?

A charge of +80 μC is placed on the x axis at x = 0.A second charge of −50 μC is placed on the x axis at x = 50 cm.What is the magnitude of the electrostatic force on a third charge of 4.0 μC placed on the x axis at x = 30 cm?

A point charge Q is placed on the x axis at x = 2.0 m.A second point charge,−Q,is placed at x = 3.0 m.If Q = 40 μC,what is the magnitude of the electrostatic force on a 30-μC charge placed at the origin?

Three identical point charges Q are placed at the vertices of an equilateral triangle (length of each side = 2.0 m) .If Q = 60 μC,what is the magnitude of the electrostatic force on any one of the charges?

A particle (charge = +40 μC) is located on the x axis at the point x = −20 cm,and a second particle (charge = −50 μC) is placed on the x axis at x = +30 cm.What is the magnitude of the total electrostatic force on a third particle (charge = −4.0 μC) placed at the origin (x = 0) ?

Three point charges,two positive and one negative,each having a magnitude of 20 μC are placed at the vertices of an equilateral triangle (30 cm on a side) .What is the magnitude of the electrostatic force on one of the positive charges?

Three point charges are positioned on the x axis.If the charges and corresponding positions are +32 μC at x = 0,+20 μC at x = 40 cm,and −60 μC at x = 60 cm,what is the magnitude of the electrostatic force on the +32-μC charge?

If a = 3.0 mm,b = 4.0 mm,Q₁ = 60 nC,Q₂ = 80 nC,and q = 32 nC in the figure,what is the magnitude of the electric force on q?

If a = 3.0 mm,b = 4.0 mm,Q₁ = −60 nC,Q₂ = 80 nC,and q = 30 nC in the figure,what is the magnitude of the electric force on q?

Each of two small non-conducting spheres is charged positively,the combined charge being 40 μC.When the two spheres are 50 cm apart,each sphere is repelled from the other by a force of magnitude 2.0 N.Determine the magnitude of the smaller of the two charges.

A point charge Q is placed on the x axis at the origin.An identical point charge is placed on the x axis at x = −1.0 m and another at x = +1.0 m.If Q = 40 μC,what is the magnitude of the electrostatic force on the charge at x = +1.0 m?