Question 1

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 2

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 always perpendicular outward and away from the charge. This is because positive charges repel other positive charges and therefore the electric field generated will be directed away from the charge in all directions.

Question 3

We have a hollow metallic sphere with charge −5.9 µC and radius 5 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

Due to the conductive property of the sphere, charges will move freely within it and distribute themselves uniformly on the surface. Thus, initially, the -5.9 µC charge is uniformly distributed on the entire surface of the sphere.

When we insert a +10 µC charge at the center of the sphere, the negative charge on the surface of the sphere is attracted toward the center of the sphere, while the positive charge on the surface is repelled outward. This redistributes the charges on the surface until a new equilibrium is reached.

To find the final charge on the surface of the sphere, we can use the fact that the net charge inside the sphere remains the same both before and after the insertion of the +10 µC charge. Initially, the net charge inside the sphere is -5.9 µC. After the +10 µC charge is inserted, the net charge inside the sphere becomes +4.1 µC (10 µC - 5.9 µC).

Therefore, the charge on the outer surface of the sphere must be equal in magnitude but opposite in sign to the net charge inside the sphere, giving us a final charge of +4.1 µC on the outer surface.

Hence, the best choice is A) +4.1 µC.

Question 4

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 5

We have an initially uncharged hollow metallic sphere with radius of 5.0 cm.I place a small object with a charge of +26 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 × 10⁹ N⋅m²/C²)

Accepted Answer

To find the electric field at a point outside the sphere, we can use the formula for the electric field of a point charge:

E = k * q / r^2

Where k is the Coulomb constant, q is the charge on the small object, and r is the distance between the small object and the point where we want to find the electric field.

In this case, the distance from the center of the sphere to the point 10 cm away is 15 cm (since the radius of the sphere is 5 cm and the small object is at the center). The charge q is given as +26 µC. Plugging in these values, we get:

E = (8.99 × 10^9 N⋅m^2/C^2) * (26 × 10^-6 C) / (0.15 m)^2
E ≈ 23.44 × 10^6 N/C

Converting to scientific notation, we get:

E ≈ 23.4 × 10^6 N/C

Therefore, the best choice is C.

Question 6

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. This is due to the fact that the charges on the surface of a conductor arrange themselves in such a way that the electric field inside the conductor is zero. As the shape of the conductor changes, the charges rearrange themselves accordingly to maintain a zero electric field inside the conductor.

Question 7

An initially uncharged hollow metallic sphere with radius of 6 cm has a small object with a charge of +19 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 metallic hollow sphere is conductive, the charges will redistribute themselves to achieve electrostatic equilibrium. Due to the presence of the small object with a positive charge (+19 µC), negative charges will move towards the surface of the sphere closest to the center of the sphere while positive charges will move to the outer surface of the sphere farthest away from the center of the sphere. Since there was no initial charge present in the sphere, the net charge on the sphere after the redistribution of charges will be +19 µC (equal to the charge of the small object). Therefore, the answer is choice D, +19 µC.

Question 8

The electric field in a cathode ray tube is supposed to accelerate electrons from 0 to 4.8 × 10⁷ m/s in a distance of 5.00 cm.What electric field is required? (m_e = 9.11 × 10−³¹ kg and e = 1.60 × 10−¹⁹ C)

Accepted Answer

The answer of The electric field in a cathode ray...

Question 9

Two point charges,separated by 1.1 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-μC 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 as many field lines, but they will radiate inwards. Hence, 20 field lines radiate in.

Question 10

Two identical spheres each carry a charge of −47.1 µC.The spheres are separated by a distance of 1.74 m.What is the electric force between the spheres? (k_e = 8.99 × 10⁹ N·m²/C²)

Accepted Answer

The electric force between two charges can be calculated using the Coulomb's Law which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = k * (q1 * q2) / d^2

where
F = electric force
k = Coulomb's constant = 8.99 × 10^9 N·m^2/C^2
q1 and q2 are the charges
d is the distance between the charges

Substituting the given values, we get

F = 8.99 × 10^9 * (-47.1*10^-6)^2 / (1.74)^2
F = 6.6 N

Since the charges are identical, they will repel each other. Therefore, the answer is (B) 6.6 N (repulsive).

Question 11

Q₁ has 50 electric field lines radiating outward and Q₂ has 150 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, it is positive, and Q2 has 150 lines converging inward, it is negative. The ratio is (50)/(−150) = −1/3.

Question 12

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

Accepted Answer

The number of electric field lines passing through a unit cross sectional area is indicative of the field strength. The closer the field lines are together, the stronger the electric field. Charge density and charge motion do not affect the spacing of the field lines. Field direction can be determined by the orientation of the field lines, but the number of field lines passing through a unit cross sectional area is not indicative of field direction.

Question 13

An electron with a speed of 3 × 10⁶ m/s moves into a uniform electric field of 600 N/C that is parallel to the electron's motion.How long does it take to bring the electron to rest? (me = 9.11 × 10−³¹ kg,e = 1.6 × 10−¹⁹ C)

Accepted Answer

The correct answer is A because the time it takes to bring the electron to rest is given by the equation: t = v/a, where v is the initial speed of the electron and a is the acceleration of the electron in the electric field. The acceleration of the electron is given by the equation: a = eE/m, where e is the charge of the electron, E is the strength of the electric field, and m is the mass of the electron. Substituting these equations into the first equation, we get: t = v/(eE/m) = (3 × 10^6 m/s)/(1.6 × 10^-19 C * 600 N/C * 9.11 × 10^-31 kg) = 2.8 × 10^-8 s.

Question 14

A proton moving at 3.0 × 10⁴ m/s is projected at an angle of 30° 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 × 10−²⁷ kg,q_proton = 1.6 × 10−¹⁹ C)

Accepted Answer

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

Question 15

In x-ray machines,electrons are subjected to electric fields as great as 6 × 10⁵ N/C.Find an electron's acceleration in this field.(m_e = 9.11 × 10−³¹ kg,e = 1.6 × 10−¹⁹ C)

Accepted Answer

The acceleration of an electron in an electric field is given by a = F/m = eE/m. Substituting the given values, a = (1.6 × 10^-19 C)(6 × 10^5 N/C) / (9.11 × 10^-31 kg) = 1.1 × 10^17 m/s².

Question 16

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

Accepted Answer

The electric field is greatest at the surface of a charged conductor. Therefore, the highest field will be on the outer surface of the hollow metallic sphere.

Question 17

Electrons in a particle beam each have a kinetic energy of 4.5 × 10−¹⁷ J.What is the magnitude of the electric field that will stop these electrons in a distance of 0.6 m? (e = 1.6 × 10−¹⁹ C)

Accepted Answer

The work done by the electric field to stop an electron with charge e and kinetic energy K is equal to the kinetic energy lost by the electron: 
W = -K = - 4.5 × 10^-19 J.
The work done by an electric field on an electron moving through a distance d is given by: 
W = eEd
where E is the magnitude of the electric field. Equating these two expressions for W and solving for E, we get: 
E = -K/(ed) = (4.5 × 10^-19)/(1.6 × 10^-19 × 0.6) = 4,687.5 N/C, which rounds to 4,688 N/C. 
Therefore, the correct answer is (C).

Question 18

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 inner surface of the sphere will have a charge of -10 µC to neutralize the +10 µC charge at the center, ensuring the electric field inside the conductor is zero.

Question 19

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 is proportional to the magnitude of the charge, so Charge C has the largest magnitude. Additionally, since Charge C has lines coming in, it must have an opposite charge to the other two. Therefore, Charge B and C will have the largest force between them.

Question 20

An airplane is flying through a thundercloud at a height of 2,500 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 +31 C at height 4,000 m within the cloud and −31 C at height 500 m,what is the magnitude of the electric field E at the aircraft? (k_e = 8.99 × 10⁹ N·m²/C²)

Accepted Answer

First, we need to find the net charge within the region where the aircraft is flying:
Q = Q_above - Q_below
Q = +31 C - (-31 C)
Q = 62 C
Then, we can use Gauss's Law to find the electric field:
∮E⃗ ⋅dA⃗ = Q/ε0
E(4πr^2) = Q/ε0
E = Q/(4πε0r^2)
where r is the distance from the charge concentration to the aircraft. Since the charges are distributed uniformly, we can assume that r is the average distance between the aircraft and the two charge concentrations, which is:
r = (4000 m + 500 m)/2 - 2500 m
r = 1000 m
Plugging in the values, we get:
E = (62 C)/(4π(8.99 × 10^9 N·m^2/C^2)(1000 m)^2)
E ≈ 193,535 N/C, which is answer choice D.

Quiz 15: Electric Forces and Electric Fields

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

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

We have a hollow metallic sphere with charge −5.9 µC and radius 5 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? ​

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

We have an initially uncharged hollow metallic sphere with radius of 5.0 cm.I place a small object with a charge of +26 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.(ke = 8.99 × 109 N⋅m2/C2) ​

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

An initially uncharged hollow metallic sphere with radius of 6 cm has a small object with a charge of +19 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? ​

The electric field in a cathode ray tube is supposed to accelerate electrons from 0 to 4.8 × 107 m/s in a distance of 5.00 cm.What electric field is required? (me = 9.11 × 10−31 kg and e = 1.60 × 10−19 C) ​

Two point charges,separated by 1.1 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-μC charge with respect to field lines? ​

Two identical spheres each carry a charge of −47.1 µC.The spheres are separated by a distance of 1.74 m.What is the electric force between the spheres? (ke = 8.99 × 109 N·m2/C2) ​

Q1 has 50 electric field lines radiating outward and Q2 has 150 field lines converging inward.What is the ratio Q1/Q2? ​

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

An electron with a speed of 3 × 106 m/s moves into a uniform electric field of 600 N/C that is parallel to the electron's motion.How long does it take to bring the electron to rest? (me = 9.11 × 10−31 kg,e = 1.6 × 10−19 C)

A proton moving at 3.0 × 104 m/s is projected at an angle of 30° 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.mproton = 1.67 × 10−27 kg,qproton = 1.6 × 10−19 C)

In x-ray machines,electrons are subjected to electric fields as great as 6 × 105 N/C.Find an electron's acceleration in this field.(me = 9.11 × 10−31 kg,e = 1.6 × 10−19 C)