Question 1

Two identical iron spheres have equal positive charges and the force between them when they are 1 m apart is 1 N.What percentage of the electrons has been removed from each sphere if each sphere has 1 mole (6 x 10²³)of iron atoms? (k_e = 8.99 x 10⁹N*m²/C²,the atomic number of iron is 26,and the electron charge is -1.6 x 10^-19 C)

Accepted Answer

The force between the spheres is given by Coulomb's law:$$F = k\frac{q_1q_2}{r^2}$$where k is Coulomb's constant, q1 and q2 are the charges on the spheres, and r is the distance between them.We are given that F = 1 N, r = 1 m, and k = 8.99 x 10^9 N F1F1F1S1F1F1F10m²/C².Substituting these values into Coulomb's law, we get:$$1 N = 8.99 x 10^9 N F1F1F1S1F1F1F10m²/C² \frac{q_1q_2}{(1 m)^2}$$$$q_1q_2 = 1.11 x 10^{-10} C^2$$Since the spheres have equal charges, we can write:$$q_1 = q_2 = \sqrt{1.11 x 10^{-10} C^2} = 1.05 x 10^{-5} C$$The number of electrons removed from each sphere is:$$n = \frac{q}{e} = \frac{1.05 x 10^{-5} C}{-1.6 x 10^{-19} C} = 6.56 x 10^{13}$$The percentage of electrons removed from each sphere is:$$\% = \frac{n}{N} imes 100\% = \frac{6.56 x 10^{13}}{6 x 10^{23}} imes 100\% = 4 x 10^{-10} \%$$Therefore, the correct answer is D.

Question 2

An airplane is flying through a thundercloud at a height of 2 000 m.(This is a very dangerous thing to do because of updrafts,turbulence,and the possibility of electric discharge. )If there is a charge concentration of +40 C at height 3 000 m within the cloud and -40 C at height 1 000 m,what is the magnitude of the electric field E at the aircraft? (k_e = 8.99 x 10⁹ N·m²/C²)

Accepted Answer

The answer of An airplane is flying through a thundercloud...

Question 3

An initially uncharged hollow metallic sphere with radius of 5 cm has a small object with a charge of +10 µC carefully placed at the center of the sphere through a hole in the latter's surface.With the charge in place,what charge is now present on the outside surface of the sphere?

Accepted Answer

Since the sphere is metallic, initially all the charges will be distributed uniformly on its surface due to electrostatic repulsion. When the small charged object is placed at the center of the sphere, the electrons in the sphere will redistribute themselves in such a way as to cancel the electric field inside the sphere. This means that the charges will move as far away from the central charge as possible. Due to the spherical symmetry of the problem, the charges will distribute themselves uniformly on the outer surface of the sphere. Since the object has a positive charge, the electrons in the sphere will move towards it, leaving a net positive charge on the outer surface of the sphere. The sphere will now have a total charge of +10 µC on the outside surface. Thus, the answer is D.

Question 4

The electric field in a cathode ray tube is supposed to accelerate electrons from 0 to 1.60 x 10⁷m/s in a distance of 2.00 cm.What electric field is required? (m_e = 9.11 x 10^-31kg and e = 1.60 x 10^-19 C)

Accepted Answer

We can use the equation:

acceleration = (electric field) x (charge of electron) / (mass of electron)

We are given the final velocity and distance, so we can find the acceleration:

a = v^2 / (2d) = (1.6 x 10^7 m/s)^2 / (2 x 0.02 m) = 1.28 x 10^12 m/s^2

Plugging in values, we get:

1.28 x 10^12 m/s^2 = (electric field) x (1.6 x 10^-19 C) / (9.11 x 10^-31 kg)

Solving for electric field, we get:

electric field = 36,400 N/C

Therefore, the best choice is C.

Question 5

Relative distribution of charge density on the surface of a conducting solid depends on:

Accepted Answer

The relative distribution of charge density on the surface of a conducting solid depends on the shape of the conductor because the charges on the surface will redistribute themselves in such a way that the electric potential is constant everywhere on the surface. This is known as the principle of equipotential surfaces. The shape of the conductor determines the shape of these equipotential surfaces and hence determines the relative distribution of charge density on the surface.

Question 6

Two identical spheres each carry a charge of -40.0 µC.The spheres are separated by a distance of 1.00 m.What is the electric force between the spheres? (k_e = 8.99 x 10⁹ N·m²/C²)

Accepted Answer

Using Coulomb's law, the electric force between the spheres can be calculated:
$$F=\frac{kq_1q_2}{r^2}$$
where k is the Coulomb constant, q1 and q2 are the charges on the spheres, and r is the distance between the centers of the spheres. 
Plugging in the given values, we get:
$$F=\frac{(8.99	imes 10^9 N\cdot m^2/C^2)(-40.0	imes10^{-6} C)(-40.0	imes10^{-6} C)}{(1.00m)^2}$$
Simplifying this expression, we get:
$$F=14.4N$$
Since both spheres carry the same charge, they repel each other and the force is repulsive. Therefore, the best choice is B, 14.4 N (repulsive).

Question 7

The electric field associated with a uniformly charged hollow metallic sphere is the greatest at:

Accepted Answer

The electric field associated with a uniformly charged hollow metallic sphere is the greatest at the sphere's outer surface because there is no electric field inside the hollow sphere due to the charge distribution. Therefore, the electric field is strongest at the outer surface where the charge density is the highest.

Question 8

The number of electric field lines passing through a unit cross sectional area is indicative of:

Accepted Answer

The electric field lines represent the direction and strength of the electric field. The density of the field lines represents the strength of the electric field. Therefore, the number of electric field lines passing through a unit cross sectional area is indicative of the electric field strength.

Question 9

The electric field at the surface of a positively charged conductor has a direction characterized by which of the following?

Accepted Answer

The electric field at the surface of a positively charged conductor is perpendicular outward and away from the charge. This is because at the surface, the electric charge distributes itself in such a way that the electric field is perpendicular to the surface. Additionally, since the conductor is positively charged, the electric field lines start at the surface and extend outward into space.

Question 10

At what point is the charge per unit area greatest on the surface of an irregularly shaped conducting solid?

Accepted Answer

The charge per unit area (surface charge density) on an irregularly shaped conducting solid is greatest where the curvature is greatest. This is because electric field lines are more concentrated in areas of high curvature, leading to a higher accumulation of charge in these regions.

Question 11

We have an initially uncharged hollow metallic sphere with radius of 5.0 cm.I place a small object with a charge of +10 µC at the center of the sphere through a hole in the surface.Find the electric field present at a point 10 cm from the sphere's center.(k_e = 8.99 x 10⁹ N*m²/C²)

Accepted Answer

The electric field at a point outside a charged sphere is given by E = k_e * Q / r^2. Here, Q = 10 µC = 10 x 10^-6 C, r = 0.1 m, and k_e = 8.99 x 10^9 N m^2/C^2. Plugging in these values gives E = 8.99 x 10^9 * 10 x 10^-6 / (0.1)^2 = 8.99 x 10^6 N/C, which is approximately 9.0 x 10^6 N/C.

Question 12

A proton moving at 3.0 x 10⁴ m/s is projected at an angle of 30 $\degree$ above a horizontal plane.If an electric field of 400 N/C is acting down,how long does it take the proton to return to the horizontal plane? (Hint: Ignore gravity.m_proton = 1.67 x 10^-27 kg,q_proton = 1.6 x 10^-19 C)

Accepted Answer

The answer of A proton moving at 3.0 x 10⁴...

Question 13

Charge A has 10 electric field lines coming out,Charge B has 20 lines coming out,and Charge C has 30 lines coming in.Which pair of these charges will have the largest force between them if placed one cm apart?

Accepted Answer

The number of electric field lines represents the strength of the electric field, and the strength of the electric field is directly proportional to the force experienced by a charge. Charge B has the strongest electric field, therefore it will experience the largest force when placed one cm apart from Charge C.

Question 14

The electric field of a point charge has an inverse ________ behavior.

Accepted Answer

The electric field of a point charge follows an inverse square law, meaning it decreases in strength as the square of the distance from the charge, which is represented by $r^{-2}$.

Question 15

Two point charges,separated by 1.5 cm,have charge values of +2.0 and -4.0 µC,respectively.Suppose we determine that 10 field lines radiate out from the +2.0-µC charge.If so,what might be inferred about the -4.0-mC charge with respect to field lines?

Accepted Answer

The number of field lines is proportional to the magnitude of the charge. Since the -4.0 µC charge is twice the magnitude of the +2.0 µC charge, it will have twice the number of field lines, but they will radiate inwards because it is a negative charge.

Question 16

Q₁ has 50 electric field lines radiating outward and Q₂ has 100 field lines converging inward.What is the ratio Q₁/Q₂?

Accepted Answer

The number of electric field lines is proportional to the charge. Since Q1 has 50 lines radiating outward (positive charge) and Q2 has 100 lines converging inward (negative charge), the ratio Q1/Q2 is 50/-100 = -1/2.

Question 17

A ping-pong ball covered with a conducting graphite coating has a mass of 5.0 x 10^-3 kg and a charge of 4.0 µC.What electric field directed upward will exactly balance the weight of the ball? (g = 9.8 m/s²)

Accepted Answer

The electric field (E) required to balance the weight of the ball can be found using the formula E = mg/q, where m is the mass, g is the acceleration due to gravity, and q is the charge. Substituting the given values: E = (5.0 x 10^-3 kg * 9.8 m/s^2) / (4.0 x 10^-6 C) = 1.2 x 10^4 N/C.

Question 18

We have a hollow metallic sphere with charge -5.0 µC and radius 5.0 cm.We insert a +10-µC charge at the center of the sphere through a hole in the surface.What charge now rests on the outer surface of the sphere?

Accepted Answer

The charge inside the sphere induces an equal and opposite charge on the inner surface of the sphere, leaving a net charge of +5 µC on the outer surface.

Question 19

An initially uncharged hollow metallic sphere with radius of 5 cm has a small object with a charge of +10 µC carefully placed at the center of the sphere through a hole in the latter's surface.What charge resides inner surface of the sphere?

Accepted Answer

The charge on the inner surface of the sphere will be -10 µC to ensure that the electric field inside the conductor is zero, as per electrostatic shielding.

Question 20

Electrons in a particle beam each have a kinetic energy of 3.2 x 10^-17 J.What is the magnitude of the electric field that will stop these electrons in a distance of 0.1 m? (e = 1.6 x 10^-19 C)

Accepted Answer

The kinetic energy of each electron is given by 1/2mv², where m is the mass of the electron and v is its velocity. Since the mass of an electron is constant, we can equate the kinetic energy to the work done by an external electric field to bring the electron to a stop:
1/2mv² = eEd
where E is the magnitude of the electric field and d is the distance traveled by the electron before coming to a stop. Solving for E, we have:
E = 1/ed * (1/2mv²/e)
Substituting the given values and solving, we get:
E = 2.0 x 10³ N/C
Therefore, the answer is choice C.

Quiz 15: Electric Forces and Electric Fields