Question 1

A thin metallic spherical shell of radius 50 cm has a total charge of 4.0 µC placed on it. At the center of the shell is placed a point charge of 2.00 µC. What is the electric field at a distance of 30.0 cm from the center of the spherical shell?

Accepted Answer

According to Gauss's Law, the electric field at a distance outside of a charged spherical shell is the same as if all of the charge were concentrated at the center of the shell. Therefore, the electric field at a distance of 30.0 cm from the center of the spherical shell is the same as the electric field created by a point charge of 2.00 µC at the center of a sphere with a radius of 30.0 cm. Using Coulomb's Law, we can determine that the magnitude of the electric field is:
E = k*(q/r^2) = (9.0 x 10^9 N*m^2/C^2)*(2.00 x 10^-6 C)/(0.30 m)^2 = 2.0 x 10^10 N/C
The direction of the electric field is outward from the center of the sphere, since the point charge is positive. Therefore, the final answer is choice B.

Question 2

An electron gun sends electrons through a region with an electric field of magnitude 1.5 × 10⁴ N/C for a distance of 2.5 cm. What is the acceleration of the electrons while in this electric field?

Accepted Answer

The answer of An electron gun sends electrons through a...

Question 3

A thin spherical shell of radius 20.0 cm has 5.0 µC of charge uniformly distributed over its surface. What is the magnitude of the electric field 2.00 m from the center of the sphere?

Accepted Answer

The answer of A thin spherical shell of radius 20.0...

Question 4

An electron is fired through the space between two oppositely charged, parallel plates. The plates form a 2.00 cm long region with an electric field of 4000 N/C perpendicular to the electron's initial velocity. If the initial speed of the electron is 1.00 × 10⁷ m/s, what angle will its new velocity make relative to the direction of the original velocity?

Accepted Answer

The answer of An electron is fired through the space...

Question 5

A thin spherical shell of radius 20.0 cm has 5.0 µC of charge uniformly distributed over its surface. What is the electric field 10.0 cm from the center of the sphere?

Accepted Answer

Inside a uniformly charged spherical shell, the electric field is zero. This is because the electric field from the charge distributed on the surface cancels out at any point inside the shell.

Question 6

What is the magnitude of the acceleration of an electron when in an electric field of magnitude 7.0 × 10⁴ N/C?

Accepted Answer

The answer of What is the magnitude of the acceleration...

Question 7

A thin metallic shell of radius 40 cm has a charge of −25 nC on it. At the center of the sphere is a point charge of 35 nC. What is the electric field 50 cm from the center of the shell?

Accepted Answer

The electric field inside a uniformly charged thin spherical shell is zero. Therefore, the electric field at 50 cm from the center of the shell is only due to the point charge at the center of the shell. Using Coulomb's law, we can find the magnitude of the electric field:
$E=\frac{kQ}{r^2}$
where $k$ is the Coulomb constant, $Q$ is the charge of the point charge, and $r$ is the distance from the point charge. Plugging in the values, we get:
$E=\frac{(9	imes10^9)(35	imes10^{-9})}{(0.5)^2}\approx 630	ext{ N/C}$
Since the point charge is positive, the electric field is outward. However, we need to take into account the negative charge on the shell, which will create an opposing electric field. The magnitude of this field can be found using the same formula, with $Q$ being the charge of the shell, which is negative:
$E=\frac{(9	imes10^9)(-25	imes10^{-9})}{(0.5)^2}\approx -450	ext{ N/C}$
Since the shell has a negative charge, the electric field it creates is inward. Therefore, the net electric field at 50 cm from the center of the shell is:
$E_{net}=630-450=180	ext{ N/C}$
Since the point charge creates a stronger electric field than the shell, the net electric field is outward. The best choice is therefore (C), 360 N/C outward.

Question 8

An infinitely long wire with a uniform linear charge distribution has a charge per unit length of 2.3 × 10^-6 C/m. What is the magnitude of the electric field at a perpendicular distance of 0.50 m from the wire?

Accepted Answer

The electric field due to an infinitely long wire with linear charge density λ at a distance r is given by E = (λ / (2πε₀r)). Substituting λ = 2.3 × 10⁻⁶ C/m, r = 0.50 m, and ε₀ = 8.85 × 10⁻¹² C²/(N·m²), we find E ≈ 8.3 × 10⁴ N/C.

Question 9

For a conductor in electrostatic equilibrium, which of the following statements is NOT true?

Accepted Answer

In electrostatic equilibrium, the charge on a conductor distributes itself not uniformly throughout the conductor but on its surface, and it tends to accumulate at points of highest curvature.

Question 10

An electron is fired through the space between two oppositely charged, parallel plates. The plates form a 2.00 cm long region with an electric field of 4000 N/C perpendicular to the electron's initial velocity. If the initial speed of the electron is 1.00 × 10⁷ m/s, how long does it take to traverse the plates?

Accepted Answer

The time to traverse the plates depends only on the initial speed and the distance between the plates, not on the electric field, since the field acts perpendicular to the motion. Time $t = \frac{d}{v} = \frac{2.00 \times 10^{-2} \, \text{m}}{1.00 \times 10^7 \, \text{m/s}} = 2.00 \times 10^{-9} \, \text{s} = 2.00 \, \text{ns}$.

Question 11

In an electron gun, electrons are accelerated through a region with an electric field of magnitude 1.5 × 10⁴ N/C for a distance of 2.5 cm. If the electrons start from rest, how fast are they moving after traversing the gun?

Accepted Answer

The speed of the electrons can be calculated using the work-energy principle. The work done on the electrons by the electric field is equal to the change in kinetic energy. Using the formula $ v = \sqrt{2 \cdot e \cdot E \cdot d / m_e} $, where $ e $ is the charge of an electron, $ E $ is the electric field, $ d $ is the distance, and $ m_e $ is the mass of an electron, the speed is approximately $ 1.1 \times 10^7 $ m/s.

Question 12

In an electron gun, electrons are accelerated through a region with an electric field of magnitude 1.5 × 10⁴ N/C for a distance of 2.5 cm. If the electrons start from rest, how long does it take for the electrons to traverse the gun?

Accepted Answer

The answer of In an electron gun, electrons are accelerated...

Question 13

If more field lines leave a closed surface than enter that surface, then

Accepted Answer

According to Gauss's law, the total electric flux through a closed surface is proportional to the net charge contained inside that surface. If more field lines leave the surface than enter, then the flux through the surface is positive and there must be a net positive charge inside the surface.

Question 14

A thin metallic spherical shell of radius 50 cm has a total charge of 4.0 µC placed on it. At the center of the shell is placed a point charge of 2.00 µC. What is the electric field at a distance of 90 cm from the center of the spherical shell?

Accepted Answer

The answer of A thin metallic spherical shell of radius...

Question 15

A conducting sphere of radius 10.0 cm holds a net charge of 4.4 µC. What is the surface charge density?

Accepted Answer

The surface charge density ($\sigma$) is calculated by dividing the total charge ($Q$) by the surface area ($A$) of the sphere. The formula for the surface area of a sphere is $4\pi r^2$, where $r$ is the radius. Given $Q = 4.4 \times 10^{-6}$ C and $r = 0.1$ m, the surface charge density is $\sigma = \frac{Q}{4\pi r^2} = \frac{4.4 \times 10^{-6} C}{4\pi (0.1)^2 m^2} = 3.5 \times 10^{-5} C/m^2$.

Question 16

A thin metallic shell of radius 40 cm has a charge of −25 nC on it. At the center of the sphere is a point charge of 35 nC. What is the electric field 15 cm from the center of the shell?

Accepted Answer

The answer of A thin metallic shell of radius 40...

Question 17

An electron is fired through the space between two oppositely charged, parallel plates. The plates form a 2.00 cm long region with an electric field of 4000 N/C perpendicular to the electron's initial velocity. What is the change in velocity of the electron after it passes through the plates?

Accepted Answer

The force on an electron in an electric field is given by $F=qE$, where $q$ is the charge of the electron and $E$ is the electric field strength. Since the electric field is perpendicular to the electron's initial velocity, the force causes the electron to undergo uniform acceleration, with $a=\frac{F}{m_e}$, where $m_e$ is the mass of the electron. The displacement of the electron is $d=0.02$ m. Using the kinematic equation $v_f^2=v_i^2+2ad$, where $v_i$ is the electron's initial velocity and $v_f$ is its final velocity, we can solve for the change in velocity: $v_f^2=v_i^2+2(qE/m_e)d$ $v_f^2=v_i^2+2(\frac{-e \cdot 4000 N/C}{9.11 imes 10^{-31} kg})(0.02 m)$ $v_f^2=v_i^2-1.756 imes 10^{14} m^2/s^2$ The change in velocity is equal to $\sqrt{v_f^2-v_i^2}$: $\Delta v = \sqrt{(v_i^2-1.756 imes 10^{14} m^2/s^2)-v_i^2}$ $\Delta v = \sqrt{-1.756 imes 10^{14} m^2/s^2}$ $\Delta v = \pm 1.325 imes 10^7 m/s$ Since the electron is negatively charged and moving in the opposite direction of the electric field, its velocity decreases. Therefore, we take the negative square root: $\Delta v = -1.325 imes 10^7 m/s \approx -1.4 imes 10^{16} m/s$ The closest answer choice is (D) 1.4 × 10⁶ m/s.

Question 18

A long wire with uniform linear charge density 1.35 nC/m is surrounded by a concentric cylindrical surface of length 1.60 m and radius 23.6 cm. What is the magnitude of the electric field at the curved surface of the cylinder?

Accepted Answer

The answer of A long wire with uniform linear charge...

Question 19

A proton is moving in an electric field. The direction of the acceleration of the proton is

Accepted Answer

The acceleration of a charged particle in an electric field is directly proportional to the strength of the electric field and has the same direction as the electric field. Therefore, the direction of the acceleration of the proton is in the direction of the electric field.

Question 20

A 1.65 nC charge experiences an acceleration of 6.33 × 10⁷ m/s² in an electric field. What is the magnitude of the electric field?

Accepted Answer

The answer of A 1.65 nC charge experiences an acceleration...

Quiz 16: Electric Forces and Fields