Question 1

A helium nucleus is located between the plates of a parallel-plate capacitor as shown. The nucleus has a charge of +2e and a mass of 6.6 × 10^-²⁷kg. What is the magnitude of the electric field such that the electric force exactly balances the weight of the helium nucleus so that it remains stationary?

Accepted Answer

The answer of A helium nucleus is located between the...

Question 2

A rigid electric dipole is free to move in the electric field represented in the figure. Which one of the following phrases most accurately describes the initial motion of the dipole if it is released from rest in the position shown?

Accepted Answer

The answer of A rigid electric dipole is free to...

Question 3

Which one of the following statements is true concerning the spacing of the electric field lines in the vicinity of two point charges of equal magnitude and opposite sign?

Accepted Answer

The spacing of the electric field lines indicates the relative magnitude of the electric field. When the charges are of equal magnitude and opposite sign, the electric field lines will be closer together near the charges and farther apart as they move away. This indicates that the electric field is strongest near the charges and weaker as you move away from them. The spacing of the electric field lines does not depend on the magnitude of the charges but instead reflects the relative strength of the electric field. Therefore, option (D) is the correct choice.

Question 4

Two point charges, A and B, lie along a line separated by a distance L. The point x is the midpoint of their separation. 
-Which combination of charges would yield the greatest repulsive force between the charges?

Accepted Answer

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Question 5

What is the magnitude and direction of the electric force on a -3.0 µC charge at a point where the electric field is 2800 N/C and is directed along the +y axis.

Accepted Answer

The electric force (F) on a charge (q) is calculated using the formula F=qE, where E is the electric field. 
In this case, the charge is -3.0 µC, which is negative, and the electric field is 2800 N/C along the +y axis, which means it is also in the -y direction for the negative charge. 
Thus, we have F = (-3.0 × 10^-6 C) × (2800 N/C) = -8.4 × 10^-3 N in the -y direction. 
Note that the magnitude of the force is always positive, so we take the absolute value before reporting the answer. 
Therefore, the magnitude of the electric force on the charge is 8.4 × 10^-3 N in the -y direction. The closest choice is (C) with 0.0084 N in the -y direction.

Question 6

A total charge of -6.50 µC is uniformly distributed within a sphere that has a radius of 0.150 m. What is the magnitude and direction of the electric field at 0.300 m from the surface of the sphere?

Accepted Answer

The total distance from the center is 0.450 m (0.150 m + 0.300 m). Using Gauss's law, the electric field E = k|Q|/r², where k = 8.99 × 10⁹ N·m²/C², Q = -6.50 µC, and r = 0.450 m. The field is radially inward due to the negative charge.

Question 7

The magnitude of the electric field at a distance of two meters from a negative point charge is E. What is the magnitude of the electric field at the same location if the magnitude of the charge is doubled?

Accepted Answer

According to Coulomb's law, the magnitude of electric field E is directly proportional to the magnitude of the charge Q and inversely proportional to the square of the distance r, i.e., E = kQ/r^2, where k is a constant. If the magnitude of the charge is doubled, the new magnitude of charge becomes 2Q. Therefore, the new magnitude of electric field at the same distance of two meters becomes E' = k(2Q)/r^2 = 2(kQ/r^2) = 2E. Hence, the correct option is D) 2E.

Question 8

Two point charges, A and B, lie along a line separated by a distance L. The point x is the midpoint of their separation. 
-Which combination of charges will yield zero electric field at the point x?

Accepted Answer

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Question 9

An electric dipole is released from rest in a uniform electric field with the orientation shown. Which entry in the table below correctly describes the net torque and the net force on the dipole?

Accepted Answer

The electric dipole experiences a torque because of the orientation of the dipole moment vector relative to the electric field. The torque is given by τ = pEsinθ, where p is the dipole moment and θ is the angle between p and E. In this case, the dipole moment is perpendicular to the electric field, so the angle θ is 90° and sinθ = 1. Therefore, the torque is maximum and given by τ = pE.
 
Since the electric field is uniform, the force on the positive charge is in the direction opposite to the electric field, and the force on the negative charge is in the direction of the electric field. The magnitudes of the two forces are equal, and they form a couple that produces the torque on the dipole. The net force on the dipole is zero, since the forces on the two charges are equal and opposite.

Therefore, the correct choice is B, which says that the net torque is non-zero and the net force is zero.

Question 10

A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle?

Accepted Answer

The electric flux (Φ) is given by the formula Φ = E * A * cos(θ), where E is the electric field magnitude, A is the area of the rectangle, and θ is the angle between the field and the normal to the surface. Here, E = 125,000 N/C, A = 2.50 m * 5.00 m = 12.5 m², and θ = 65.0°. Thus, Φ = 125,000 * 12.5 * cos(65°) = 6.60 × 10⁵ N·m²/C.

Question 11

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux through the loop is a maximum. At this position, the electric flux is 7.50 × 10⁵ N.m²/C. Determine the magnitude of the electric field.

Accepted Answer

The electric flux through a circular loop of radius r in a uniform electric field E is given by:

Φ = πr²E

Since the diameter of the loop is given as 0.626 m, the radius is 0.313 m.

At the position where the electric flux is a maximum, the loop is perpendicular to the electric field. Therefore, we can equate the given electric flux to the above equation to find the magnitude of the electric field:

7.50 × 10⁻¹⁵ C/N = π(0.313 m)²E
E = (7.50 × 10⁻¹⁵ C/N) / π(0.313 m)²
E = 2.44 × 10¹⁶ N/C

Therefore, the answer is (C).

Question 12

The figure shows the electric field lines in the vicinity of two point charges. Which one of the following statements concerning this situation is true?

Accepted Answer

The answer of The figure shows the electric field lines...

Question 13

A cubical Gaussian surface is placed in a uniform electric field as shown in the figure. The length of each edge of the cube is 1.0 m. The uniform electric field has a magnitude of 5.0 × 10⁸ N/C and passes through the left and right sides of the cube perpendicular to the surface. What is the total electric flux that passes through the cubical Gaussian surface?

Accepted Answer

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Question 14

Two identical conducting spheres carry charges of +5.0 $\mu$C and -1.0 $\mu$C, respectively. The centers of the spheres are initially separated by a distance L. The two spheres are brought together so that they are in contact. The spheres are then returned to their original separation L. What is the ratio of the magnitude of the electric force on either sphere after the spheres are touched to that before they were touched?

Accepted Answer

Before the spheres are touched, the magnitude of the electric force on each sphere is given by:

$F_1 = F_2 = \frac{kq^2}{L^2}$

After the spheres are touched, they will have the same charge of $Q=5-1=4\,\mu C$. The total charge is conserved, so each sphere will have a final charge of $2\,\mu C$.

The electric force between the spheres after they are touched is given by:

$F' = \frac{kq^2}{4L^2}$

The ratio of the electric forces before and after touching is:

$\frac{F'}{F_1} = \frac{\frac{kq^2}{4L^2}}{\frac{kq^2}{L^2}} = \frac{1}{4}$

Similarly, the ratio of the electric forces after touching and before touching is:

$\frac{F_2}{F'} = \frac{\frac{kq^2}{L^2}}{\frac{kq^2}{4L^2}} = \frac{4}{1}$

Multiplying these ratios, we get:

$\frac{F_2}{F_1} = \frac{F_2}{F'} \cdot \frac{F'}{F_1} = \frac{4}{1} \cdot \frac{1}{4} = \boxed{	extbf{(B) } \frac{4}{5}}$

Question 15

A conducting sphere carries a net charge of +6 µC. The sphere is located at the center of a conducting spherical shell that carries a net charge of -2 µC. Determine the excess charge on the outer surface of the spherical shell.

Accepted Answer

The answer of A conducting sphere carries a net charge...

Question 16

A solid, conducting sphere of radius a carries an excess charge of +6 µC. This sphere is located at the center of a hollow, conducting sphere with an inner radius of b and an outer radius of c as shown. The hollow sphere also carries a total excess charge of +6 µC. 
-Determine the excess charge on the inner surface of the outer sphere (a distance b from the center of the system).

Accepted Answer

The excess charge on the inner surface of the outer sphere is equal in magnitude and opposite in sign to the charge on the inner sphere, which is -6 $\mu$C, due to electrostatic induction.

Question 17

A solid, conducting sphere of radius a carries an excess charge of +6 µC. This sphere is located at the center of a hollow, conducting sphere with an inner radius of b and an outer radius of c as shown. The hollow sphere also carries a total excess charge of +6 µC. 
-Determine the excess charge on the outer surface of the outer sphere (a distance c from the center of the system).

Accepted Answer

The answer of A solid, conducting sphere of radius a...

Question 18

Which one of the following statements is true concerning the electrostatic charge on a conductor?

Accepted Answer

In a conductor, electrostatic charges reside entirely on the surface and distribute themselves according to the shape of the object to minimize repulsion.

Question 19

What is the electric flux passing through a Gaussian surface that surrounds a +0.075 C point charge?

Accepted Answer

The electric flux passing through a Gaussian surface surrounding a point charge is given by q/ε0 where q is the charge enclosed by the surface and ε0 is the electric constant. Plugging in the given values, we get (0.075 C)/(8.85×10^-12 N^-1m^-2C^-2) = 8.5×10^10 Nm^2C^-1, which is closest to option A.

Question 20

A straight, copper wire has a length of 0.50 m and an excess charge of -1.0 × 10^-⁵C distributed uniformly along its length. Find the magnitude of the electric field at a point located 7.5 × 10^-³ m from the midpoint of the wire.

Accepted Answer

The answer of A straight, copper wire has a length...

Quiz 18: Electric Forces and Electric Fields