Question 1

FIGURE 20-9 
-Two point charges of +2.00 μC and +4.00 μC are placed at the origin and at y = -0.300 m, as shown in Figure 20-9. What is the electric field potential at a point P at a position x = 0.400 m due to these charges?

Accepted Answer

The answer of FIGURE 20-9 
-Two point charges of +2.00...

Question 2

A 4.0-μC charge is situated at the origin of an xy-coordinate system. What is the potential difference between a point x₁ = 2.0 m and another point x₂ = 5.0 m because of this charge?

Accepted Answer

The potential difference V between two points due to a point charge is given by V = kQ(1/r1 - 1/r2), where k = 8.99 × 10^9 N·m²/C², Q = 4.0 × 10^-6 C, r1 = 2.0 m, and r2 = 5.0 m. Calculating gives V = 11 × 10^3 V.

Question 3

FIGURE 20-10 
-Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential at point B due to these two charges?

Accepted Answer

The potential due to point charges can be calculated using the formula:

V = kq/r

where V is the potential, k is the Coulomb constant, q is the charge, and r is the distance between the point charge and the point where potential is to be calculated.

For point B, the potential due to the +4.00 μC charge is:

V1 = kq1/r1 = (9 × 10^9 N m^2/C^2)(4.00 × 10^-6 C)/(1.00 m) = 3.60 × 10^3 V

For point B, the potential due to the +2.00 μC charge is:

V2 = kq2/r2 = (9 × 10^9 N m^2/C^2)(2.00 × 10^-6 C)/(√2 m) = 1.27 × 10^3 V

The total potential at point B is the sum of the potentials due to the two charges:

V = V1 + V2 = 3.60 × 10^3 V + 1.27 × 10^3 V = 4.87 × 10^3 V

Converting to scientific notation, we get:

V = 4.87 × 10^3 V = 4.87 × 10^3 × 10^-6 kV = 4.87 × 10^-3 kV = 4.87 mV

Finally, since the question asks for the potential in volts, we need to convert millivolts to volts:

V = 4.87 mV = 4.87 × 10^-3 V

The closest answer choice to this value is E) 113 × 10^-3 V, which is equivalent to 113 mV, so the answer is E).

Question 4

FIGURE 20-7 
-Two identical charges of magnitude 6.0 μC are placed at the corners of the base of an equilateral triangle, as shown in Figure 20-7. What is the electric potential at the vertex, P, of the triangle due to these charges?

Accepted Answer

The electric potential at a point due to a point charge is given by V = kQ/r. For two identical charges, the total potential is the sum of potentials due to each charge. Here, V = 2(k * 6.0 μC / r). Assuming r is the side of the equilateral triangle, the calculation gives 54 kV.

Question 5

A 4.0-μC charge is situated at the origin of an xy-coordinate system. What is the potential difference between a point x = 4.0 m and y = -4.0 m because of this charge?

Accepted Answer

The potential difference between two points at the same distance from a point charge is zero because the electric potential depends only on the distance from the charge, not the direction. Both points are equidistant from the charge.

Question 6

Point charges of + 2.0 μC, + 4.0 μC, and + 6.0 μC are placed at the origin, the point y = -0.30 m, and x = 0.40 m, respectively. What is the electric potential energy of the system?

Accepted Answer

The electric potential energy of a system of point charges can be calculated using the equation: 
U = (1/4πε0) ΣΣ(qi*qj/ri,j)
where U is the potential energy, qi and qj are the charges, ri,j is the distance between charges i and j, and ε0 is the permittivity of free space.

We can calculate the potential energy of the system by considering each pair of charges and summing up the contributions.

U = (1/4πε0)[(2*4/0.4)+ (2*6/√(0.4^2+0.3^2))+ (4*6/√(0.4^2+0.3^2))]

U = 0.94 mJ = 940 μJ

Therefore, the correct answer is C, 940 mJ.

Question 7

A 4.0-μC charge is situated at the origin of an xy-coordinate system. What is the magnitude and direction of the electric potential at x = 6.0 m?

Accepted Answer

The electric potential at a point a distance r away from a point charge q is given by:

V = kq/r

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

In this case, the point charge is located at the origin (r = 6.0 m), and q = 4.0 μC = 4.0 x 10^-6 C. Substituting these values into the equation gives:

V = (9 x 10^9 N m^2/C^2)(4.0 x 10^-6 C)/(6.0 m) = 6.0 x 10^3 V

Since the electric potential is a scalar quantity, it has no direction, so the answer is (E) 6000 V with no direction.

Question 8

FIGURE 20-8 
-Two charges of magnitude 6.0 μC, but opposite signs, are placed at the corners of the base of an equilateral triangle, as shown in Figure 20-8. What is the electric potential at the vertex, P, of the triangle due to these charges?

Accepted Answer

The electric potential due to a point charge is given by V = kQ/r. Since the charges are equal in magnitude but opposite in sign, their potentials at point P will cancel each other out, resulting in a net potential of 0 V.

Question 9

A +8.00-μC charge is situated along the +y-axis at y = 0.400 m. What is the electric potential at the origin because of this charge?

Accepted Answer

The electric potential at the origin due to a point charge is given by:V = kq/rwhere k is Coulomb's constant, q is the charge, and r is the distance from the charge to the point where the potential is being calculated.In this case, q = +8.00 μC = +8.00 × 10^-6 C, and r = 0.400 m. Substituting these values into the equation, we get:V = (9 × 10^9 N·m^2/C^2)(+8.00 × 10^-6 C)/(0.400 m) = +180 × 10^3 VTherefore, the electric potential at the origin due to the +8.00-μC charge is +180 × 10^3 V.

Question 10

Two point charges of magnitude 4.0 μC and -4.0 μC are situated along the x-axis at x₁ = 2.0 m and x₂ = -2.0 m, respectively. What is the electric potential at the origin of the xy-coordinate system?

Accepted Answer

The answer of Two point charges of magnitude 4.0 μC...

Question 11

A -4.0-μC charge is situated at the origin of an xy-coordinate system. Another -4.0 μC charge is at y = 4.0 m. What is the electric potential at a point x = 3.0 m due to these two charges?

Accepted Answer

The answer of A -4.0-μC charge is situated at the...

Question 12

FIGURE 20-10 
-Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential at point A due to these charges?

Accepted Answer

The answer of FIGURE 20-10 
-Two point charges of magnitude...

Question 13

Two equal point charges of magnitude 4.0 μC are situated along the x-axis at x₁ = -2.0 m and x₂ = 2.0 m. What is the electric potential at the origin of the xy-coordinate system?

Accepted Answer

To find the electric potential at the origin due to the two point charges, we use the formula:

V = kq1/r1 + kq2/r2

where k is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r1 and r2 are the distances from the charges to the origin.

Since the charges are equal and opposite, the potential due to one charge will be equal in magnitude and opposite in sign to that due to the other charge. Thus, we have:

V = kq/r1 - kq/r2

where q = 4.0 μC and r1 = 2.0 m and r2 = 2.0 m.

Plugging in the values, we get:

V = (9.0 × 10^9 N⋅m^2/C^2)(4.0 × 10^-6 C/2.0 m) - (9.0 × 10^9 N⋅m^2/C^2)(4.0 × 10^-6 C/2.0 m)

V = 36 × 10^3 V - 36 × 10^3 V

V = 0 V

Therefore, the electric potential at the origin is 0 V. So, the correct answer is choice C.

Question 14

A -4.0-μC charge is situated at the origin of an xy-coordinate system. What is the potential difference between a point x = 4.0 m and y = 2.0 m because of this charge?

Accepted Answer

The answer of A -4.0-μC charge is situated at the...

Question 15

FIGURE 20-6 
-Four equal point charges of magnitude 6.00 μC and of varying signs are placed at the corners of a square 2.00 m on each side, as shown in Figure 20-6. What is the electric potential at the center of the square?

Accepted Answer

The electric potential at the center of the square due to equal and opposite charges placed symmetrically cancels out, resulting in a net potential of 0 V.

Question 16

Two point charges q₁ = 4.0 μC and q₂ = -8.0 μC are placed along the x-axis at x₁ = 0 m and x₂ = 0.20 m, respectively. What is the electric potential energy of this system of charges?

Accepted Answer

The electric potential energy of a system of charges is given by the equation:$$U = k\frac{q_1q_2}{r}$$where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.In this case, q1 = 4.0 μC, q2 = -8.0 μC, and r = 0.20 m. Substituting these values into the equation, we get:$$U = (9.0 	imes 10^9 \frac{Nm^2}{C^2})\frac{(4.0 	imes 10^{-6} C)(-8.0 	imes 10^{-6} C)}{(0.20 m)}$$$$U = -1.44 J$$Therefore, the electric potential energy of this system of charges is -1.44 J.

Question 17

Three equal charges of magnitude + 4.00 μC are placed at the corners of an equilateral triangle of side 2.00 cm. What is the electric field potential energy of the system of these charges?

Accepted Answer

The answer of Three equal charges of magnitude + 4.00...

Question 18

The absolute value of the electric potential at a distance of 4 m from a certain point charge is 200 V. What is the absolute value of this potential at a distance of 8 m from the same point charge?

Accepted Answer

The electric potential at a distance r from a point charge q is given by V = kq/r, where k is Coulomb's constant. 
We are given V1 = 200 V at r1 = 4 m. 
Using this, we can find q = V1r1/k = (200 V)(4 m)/k. 
Now, we can find V2 at r2 = 8 m using V2 = kq/r2 = (kV1r1)/(k r2) = V1(r1/r2). 
Substituting in the given values, we get V2 = (200 V)(4 m/8 m) = 100 V. 
Therefore, the absolute value of the potential at 8 m from the point charge is 100 V. 
So, the correct answer is Choice C.

Question 19

A -4.0-μC charge is situated at the origin of an xy-coordinate system. What is the magnitude and direction of an electric potential at x = 6.0 m?

Accepted Answer

The electric potential due to a point charge is given by V = kQ/r, where k is Coulomb's constant (8.99 x 10^9 N m²/C²), Q is the charge, and r is the distance from the charge. The potential is a scalar quantity, so it has no direction. For a -4.0 μC charge at 6.0 m, V = (8.99 x 10^9)(-4.0 x 10^-6)/6 = -6000 V.

Question 20

A 4.0-μC charge is situated at the origin of an xy-coordinate system. What is the potential difference between the points x = 4.0 m and y = 2.0 m because of this charge?

Accepted Answer

The answer of A 4.0-μC charge is situated at the...

Quiz 20: Electric Potential and Electric Potential Energy

FIGURE 20-9 -Two point charges of +2.00 μC and +4.00 μC are placed at the origin and at y = -0.300 m, as shown in Figure 20-9. What is the electric field potential at a point P at a position x = 0.400 m due to these charges?

A 4.0-μC charge is situated at the origin of an xy-coordinate system. What is the potential difference between a point x1 = 2.0 m and another point x2 = 5.0 m because of this charge?

FIGURE 20-10 -Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential at point B due to these two charges?

FIGURE 20-7 -Two identical charges of magnitude 6.0 μC are placed at the corners of the base of an equilateral triangle, as shown in Figure 20-7. What is the electric potential at the vertex, P, of the triangle due to these charges?