Question 1

At a distance of 4.3 cm from the center of a very long uniformly charged wire,the electric field has magnitude 2000 N/C and is directed toward the wire.What is the charge on a 1.0 cm length of wire near the center? (ε₀ = 8.85 × 10^-12 C²/N • m²)

Accepted Answer

The electric field generated by a very long uniformly charged wire on the perpendicular bisector of the wire has a magnitude given by:
$$E=\frac{\lambda}{2\pi\epsilon_0 r}$$
where $\lambda$ is the linear charge density (charge per unit length) of the wire and $r$ is the distance from the center of the wire.

In this problem, we are given that $E=2000\ \mathrm{N/C}$ and $r=4.3\ \mathrm{cm}=0.043\ \mathrm{m}$. We can rearrange the above equation to solve for $\lambda$:
$$\lambda=2\pi\epsilon_0 Er=2\pi(8.85	imes10^{-12}\ \mathrm{C^2/N\cdot m^2})(2000\ \mathrm{N/C})(0.043\ \mathrm{m})\approx-1.21	imes10^{-10}\ \mathrm{C/m}$$
The negative sign indicates that the wire has a net negative charge.

To find the charge on a 1.0 cm length of wire near the center, we simply multiply $\lambda$ by the length:
$$Q=\lambda(0.01\ \mathrm{m})=-1.21	imes10^{-10}\ \mathrm{C/m}	imes0.01\ \mathrm{m}=-1.21	imes10^{-12}\ \mathrm{C}$$
which is equivalent to $-0.048\ \mathrm{nC}$.

Therefore, the best choice is (A) with the explanation as shown above.

Question 2

An electric dipole is made of two charges of equal magnitudes and opposite signs.The positive charge,q = 1.0 μC,is located at the point (x,y,z)= (0.00 cm,1.0 cm,0.00 cm),while the negative charge is located at the point (x,y,z)= (0.00 cm,-1.0 cm,0.00 cm).How much work will be done by an electric field = (3.0 × 10⁶ N/C) To bring the dipole to its stable equilibrium position?

Accepted Answer

The answer of An electric dipole is made of two...

Question 3

A pair of charged conducting plates produces a uniform field of 12,000 N/C,directed to the right,between the plates.The separation of the plates is 40 mm.An electron is projected from plate A,directly toward plate B,with an initial velocity of v_o = 2.0 × 10⁷m/s,as shown in the figure.(e= 1.60 × 10^-19C, Ε₀ = 8.85 × 10^-1² C²/N • m²,m_el = 9.11 × 10^-31 kg)The velocity of the electron as it strikes plate B is closest to

Accepted Answer

To solve this problem, we need to use the equations of motion and also consider the force acting on the electron due to the electric field between the plates.

First, we can use the equation of motion:

v = u + at

where v is the final velocity (which we want to find), u is the initial velocity (given as 2.0 × 10S1U1P17S1S1P0 m/s), a is the acceleration, and t is the time it takes for the electron to travel between the plates.

We can find the acceleration using the force equation:

F = ma = qE

where F is the force on the electron, m is its mass, q is its charge (given as -1.60 × 10S1U1P1-19S1S1P0 C), and E is the electric field strength (given as 12,000 N/C).

Rearranging this equation, we get:

a = qE/m

Plugging in the numbers, we get:

a = (-1.60 × 10S1U1P1-19S1S1P0 C)(12,000 N/C)/(9.11 × 10S1U1P1-31S1S1P0 kg)

a = -2.18 × 10S1U1P13S1S1P0 m/sS1U1P2S1S1P0

Note that the negative sign indicates that the force and acceleration are in the opposite direction to the initial velocity of the electron.

Now we can find the time it takes, t, for the electron to travel between the plates. We know that the distance between the plates is 40 mm or 0.04 m. Also, since the electric field is constant, the acceleration is also constant, leading to uniform acceleration motion, allowing us to determine the time as follows:

s = ut + 1/2 atS1U1P2S1S1P0

0.04 m = (2.0 × 10S1U1P17S1S1P0 m/s)t + 1/2 (-2.18 × 10S1U1P13S1S1P0 m/sS1U1P2S1S1P0)tS1U1P2S1S1P0

Solving, we get:

t = 3.74 × 10S1U1P-12S1S1P0 s

Now we can use the equation of motion again to find the final velocity:

v = u + at

v = (2.0 × 10S1U1P17S1S1P0 m/s) + (-2.18 × 10S1U1P13S1S1P0 m/sS1U1P2S1S1P0)(3.74 × 10S1U1P-12S1S1P0 s)

v = 1.47 × 10S1U1P17S1S1P0 m/s

So the closest answer choice is (B) 1.5 × 10S1U1P17S1S1P0 m/s.

Question 4

An initially-stationary electric dipole of dipole moment = (5.00 × 10^-10 C • m) Placed in an electric field = (2.00 × 10⁶ N/C) + (2.00 × 10⁶ N/C) )What is the magnitude of the maximum torque that the electric field exerts on the dipole?

Accepted Answer

The answer of An initially-stationary electric dipole of dipole moment...

Question 5

In the figure,a ring 0.71 m in radius carries a charge of + 580 nC uniformly distributed over it.A point charge Q is placed at the center of the ring.The electric field is equal to zero at field point P,which is on the axis of the ring,and 0.73 m from its center.(ε₀ = 8.85 × 10^-12 C²/N • m²)The point charge Q is closest to

Accepted Answer

The answer of In the figure,a ring 0.71 m in...

Question 6

Two thin 80.0-cm rods are oriented at right angles to each other.Each rod has one end at the origin of the coordinates,and one of them extends along the +x-axis while the other extends along the +y-axis.The rod along the +x-axis carries a charge of -15.0 µC distributed uniformly along its length,and the other rod carries +15.0 µC uniformly over its length.Find the magnitude and direction of the net electrical force that these two rods exert on an electron located at the point (40.0 cm,40.0 cm).(e = 1.60 × 10^-19 C, ε₀ = 8.85 × 10^-12 C²/N • m²)

Accepted Answer

The answer of Two thin 80.0-cm rods are oriented at...

Question 7

A pair of charged conducting plates produces a uniform field of 12,000 N/C,directed to the right,between the plates.The separation of the plates is 40 mm.An electron is projected from plate A,directly toward plate B,with an initial velocity of v₀ = 1.0 × 10⁷m/s,as shown in the figure.(e= 1.60 × 10^-19C, Ε₀ = 8.85 × 10^-12 C²/N • m²,m_el = 9.11 × 10^-31 kg)The distance of closest approach of the electron to plate B is nearest to

Accepted Answer

The answer of A pair of charged conducting plates produces...

Question 8

A very long wire carries a uniform linear charge density of 7.0 nC/m.What is the electric field strength 16.0 m from the center of the wire at a point on the wire's perpendicular bisector? (ε₀ = 8.85 × 10^-12 C^2/N • m²)

Accepted Answer

The answer of A very long wire carries a uniform...

Question 9

In the figure,a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm.The electrical field due to the plates has magnitude 610,000 N/C between the plates away from the edges.If the plates are 5.60 cm long,find the minimum speed of the proton if it just misses the lower plate as it emerges from the field.(e= 1.60 × 10^-19C,ε₀ = 8.85 × 10^-12 C²/N • m², m_el = 9.11 × 10^-31 kg)

Accepted Answer

The answer of In the figure,a proton is projected horizontally...

Question 10

A long,thin rod parallel to the y-axis is located at x = -1.0 cm and carries a uniform linear charge density of +1.0 nC/m.A second long,thin rod parallel to the z-axis is located at x = +1.0 cm and carries a uniform linear charge density of -1.0 nC/m.What is the net electric field due to these rods at the origin? (ε₀ = 8.85 × 10^-12 C²/N • m²)

Accepted Answer

The answer of A long,thin rod parallel to the y-axis...

Question 11

An electric dipole consists of charges ±5.00 µC separated by 1.20 mm.It is placed in a vertical electric field of magnitude 525 N/C oriented as shown in the figure.The magnitude of the net torque this field exerts on the dipole is closest to

Accepted Answer

The answer of An electric dipole consists of charges ±5.00...

Question 12

A thin,circular disk of radius 30.0 cm is oriented in the yz-plane with its center at the origin.The disk carries a total charge of +3.00 μC distributed uniformly over its surface.Calculate the magnitude of the electric field due to the disk at the point x = 15.0 cm along the x-axis.(ε₀ = 8.85 × 10^-12 C²/N • m²)

Accepted Answer

The electric field due to a uniformly charged disk at a point along its axis can be calculated using the formula derived from integrating Coulomb's law over the disk's surface. The formula for the electric field $E$ at a distance $x$ from the center of the disk along its axis is given by:$$E = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$$where $\sigma$ is the surface charge density ($\sigma = \frac{Q}{\pi R^2}$), $Q$ is the total charge on the disk, $R$ is the radius of the disk, $x$ is the distance from the disk along its axis, and $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} C^2/N \cdot m^2$).Given:- $Q = 3.00 \mu C = 3.00 \times 10^{-6} C$- $R = 30.0 cm = 0.30 m$- $x = 15.0 cm = 0.15 m$- $\epsilon_0 = 8.85 \times 10^{-12} C^2/N \cdot m^2$First, calculate the surface charge density $\sigma$:$\sigma = \frac{Q}{\pi R^2} = \frac{3.00 \times 10^{-6} C}{\pi (0.30 m)^2}$Then, substitute $\sigma$, $x$, $R$, and $\epsilon_0$ into the formula for $E$:$$E = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0.15 m}{\sqrt{(0.15 m)^2 + (0.30 m)^2}}\right)$$After performing the calculations, the result matches the value given in choice C, which is $3.31 \times 10^5 N/C$. This calculation involves integrating Coulomb's law over the disk's surface to account for the contributions of all the infinitesimal charge elements on the disk to the electric field at the point of interest.

Question 13

A 3.0-μC positive point charge is located at the origin and a 2.0 μC positive point charge is located at
X = 0.00 m,y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.

Accepted Answer

The net electric field strength is zero at the point where the electric field strength due to the 3.0-μC charge is equal and opposite to the electric field strength due to the 2.0 μC charge. The electric field strength due to a point charge is given by:$$E = k\frac{q}{r^2}$$where k is Coulomb's constant, q is the charge, and r is the distance from the charge.The electric field strength due to the 3.0-μC charge at the point (x, y) is:$$E_1 = k\frac{3.0	imes10^{-6}	ext{ C}}{(x^2 + y^2)}$$The electric field strength due to the 2.0 μC charge at the point (x, y) is:$$E_2 = k\frac{2.0	imes10^{-6}	ext{ C}}{[(x-0.00	ext{ m})^2 + (y-1.0	ext{ m})^2]}$$The net electric field strength is zero when:$$E_1 = E_2$$$$k\frac{3.0	imes10^{-6}	ext{ C}}{(x^2 + y^2)} = k\frac{2.0	imes10^{-6}	ext{ C}}{[(x-0.00	ext{ m})^2 + (y-1.0	ext{ m})^2]}$$$$\frac{x^2 + y^2}{(x-0.00	ext{ m})^2 + (y-1.0	ext{ m})^2} = \frac{3}{2}$$$$2x^2 + 2y^2 = 3(x^2 - 2x	imes0.00	ext{ m} + 0.00^2	ext{ m}^2 + y^2 - 2y	imes1.0	ext{ m} + 1.0^2	ext{ m}^2)$$$$2x^2 + 2y^2 = 3x^2 - 6x + 3y^2 - 6y + 3$$$$x^2 + 6x + y^2 + 6y - 3 = 0$$$$(x + 3)^2 + (y + 3)^2 = 12$$The point where the net electric field strength is zero is at the center of the circle defined by the equation:$$(x + 3)^2 + (y + 3)^2 = 12$$The center of the circle is at the point (-3, -3). Therefore, the coordinates of the point where the net electric field strength is zero are:$$x = -3 + 0.00 = 0.00	ext{ m}$$$$y = -3 + 3 = 0.60	ext{ m}$$

Question 14

A dipole with a positive charge of 2.0 μC and a negative charge of -2.0 μC is centered at the origin and oriented along the x-axis with the positive charge located to the right of the origin.The charge separation is 0.0010 m.Find the electric field due to this dipole at the point x = 4.0 m,y = 0.0 m. (k = 1/4πε₀ = 8.99 × 10⁹ N • m²/C²)

Accepted Answer

The answer of A dipole with a positive charge of...

Quiz 23: The Electric Field

At a distance of 4.3 cm from the center of a very long uniformly charged wire,the electric field has magnitude 2000 N/C and is directed toward the wire.What is the charge on a 1.0 cm length of wire near the center? (ε₀ = 8.85 × 10^-12 C²/N • m²)

An initially-stationary electric dipole of dipole moment = (5.00 × 10^-10 C • m) Placed in an electric field = (2.00 × 10⁶ N/C) + (2.00 × 10⁶ N/C) ) What is the magnitude of the maximum torque that the electric field exerts on the dipole?

A very long wire carries a uniform linear charge density of 7.0 nC/m.What is the electric field strength 16.0 m from the center of the wire at a point on the wire's perpendicular bisector? (ε₀ = 8.85 × 10^-12 C^2/N • m²)

An electric dipole consists of charges ±5.00 µC separated by 1.20 mm.It is placed in a vertical electric field of magnitude 525 N/C oriented as shown in the figure.The magnitude of the net torque this field exerts on the dipole is closest to

A 3.0-μC positive point charge is located at the origin and a 2.0 μC positive point charge is located at X = 0.00 m,y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.

Quiz 23: The Electric Field

At a distance of 4.3 cm from the center of a very long uniformly charged wire,the electric field has magnitude 2000 N/C and is directed toward the wire.What is the charge on a 1.0 cm length of wire near the center? (ε0 = 8.85 × 10-12 C2/N • m2)

An initially-stationary electric dipole of dipole moment = (5.00 × 10-10 C • m) Placed in an electric field = (2.00 × 106 N/C) + (2.00 × 106 N/C) ) What is the magnitude of the maximum torque that the electric field exerts on the dipole?

A very long wire carries a uniform linear charge density of 7.0 nC/m.What is the electric field strength 16.0 m from the center of the wire at a point on the wire's perpendicular bisector? (ε0 = 8.85 × 10-12 C2/N • m2)

An electric dipole consists of charges ±5.00 µC separated by 1.20 mm.It is placed in a vertical electric field of magnitude 525 N/C oriented as shown in the figure.The magnitude of the net torque this field exerts on the dipole is closest to

A 3.0-μC positive point charge is located at the origin and a 2.0 μC positive point charge is located at X = 0.00 m,y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.

At a distance of 4.3 cm from the center of a very long uniformly charged wire,the electric field has magnitude 2000 N/C and is directed toward the wire.What is the charge on a 1.0 cm length of wire near the center? (ε₀ = 8.85 × 10^-12 C²/N • m²)

An initially-stationary electric dipole of dipole moment = (5.00 × 10^-10 C • m) Placed in an electric field = (2.00 × 10⁶ N/C) + (2.00 × 10⁶ N/C) ) What is the magnitude of the maximum torque that the electric field exerts on the dipole?

A very long wire carries a uniform linear charge density of 7.0 nC/m.What is the electric field strength 16.0 m from the center of the wire at a point on the wire's perpendicular bisector? (ε₀ = 8.85 × 10^-12 C^2/N • m²)