Question 1

The electric field between two charged plates is 7.2 kN/C. If the plates are separated by 3.0 mm, what is the potential difference between them?

Accepted Answer

The potential difference (V) between two plates is given by V = E * d, where E is the electric field and d is the separation distance. Here, E = 7.2 kN/C = 7200 N/C and d = 3.0 mm = 0.003 m. Thus, V = 7200 * 0.003 = 21.6 V, which rounds to 22 V.

Question 2

A parallel plate capacitor, filled with air, has circular plates with a separation gap of 1.5 mm. If it is desired to store 12 mJ in this capacitor, what must the plate area be for a voltage of 1.5 kV?

Accepted Answer

The answer of A parallel plate capacitor, filled with air,...

Question 3

A pair of parallel conducting plates have potentials of −17 V and 55 V, respectively. An electron is ejected from the plate at potential −17 V and travels to the plate at 55 V. What is the speed of the electron as it strikes the 55 V plate, assuming it left the −17 V plate starting from rest? The mass of the electron is 9.11 × 10⁻³¹ kg.

Accepted Answer

The speed of the electron can be found using the principle of conservation of energy. The potential energy difference between the two plates is converted into kinetic energy as the electron moves from the lower to the higher potential. The potential difference (V) is 55 V - (-17 V) = 72 V. The charge of an electron (e) is -1.6 × 10^-19 C (the negative sign indicates the charge of the electron but does not affect the calculation of energy since it will be squared in the kinetic energy formula or taken as absolute value in potential energy calculations). The initial kinetic energy is 0 since the electron starts from rest. The final kinetic energy (K.E.) is given by K.E. = 1/2 m v^2, where m is the mass of the electron (9.11 × 10^-31 kg) and v is the velocity of the electron. The potential energy (P.E.) difference is given by P.E. = eV, where V is the potential difference.Thus, eV = 1/2 m v^2.Rearranging for v gives v = sqrt((2eV)/m).Substituting the given values: v = sqrt((2 * 1.6 × 10^-19 C * 72 V) / (9.11 × 10^-31 kg)).Calculating this gives a velocity v ≈ 5.0 × 10^6 m/s.

Question 4

A parallel plate capacitor has a dielectric with constant 3.5. The plates are circular with radius 0.07 m, and the separation gap is 0.375 mm. What energy may be stored in this capacitor if 1250 V is the applied voltage?

Accepted Answer

The energy stored in a capacitor can be calculated using the formula:

$U = \frac{1}{2}CV^2$

where U is the energy stored, C is the capacitance, and V is the voltage.

The capacitance of a parallel plate capacitor with dielectric K and separation distance d can be calculated using the formula:

$C = \frac{K\epsilon_0A}{d}$

where $\epsilon_0$ is the permittivity of free space and A is the area of the plates.

Substituting the given values, we get:

$A = \pi r^2 = \pi (0.07	ext{ m})^2 = 0.0154	ext{ m}^2$

$d = 0.375	ext{ mm} = 0.000375	ext{ m}$

$K = 3.5$

$\epsilon_0 = 8.85 	imes 10^{-12}	ext{ F/m}$

$C = \frac{K\epsilon_0A}{d} = \frac{(3.5)(8.85 	imes 10^{-12}	ext{ F/m})(0.0154	ext{ m}^2)}{0.000375	ext{ m}} = 2.263 	imes 10^{-11}	ext{ F}$

Now, we can use the formula for energy to find the answer:

$U = \frac{1}{2}CV^2 = \frac{1}{2}(2.263 	imes 10^{-11}	ext{ F})(1250	ext{ V})^2 = 9.9 	imes 10^{-4}	ext{ J} = \boxed{9.9 	imes 10^{-4}	ext{ J}}$

Question 5

A pair of electric charges has a positive potential energy. This implies that

Accepted Answer

When two electric charges have the same sign (both positive or both negative), they repel each other, which requires work to be done to bring them closer together, resulting in positive potential energy.

Question 6

The electric potential at a distance d from a point charge is V. At a distance d', the potential is 2V. What is the ratio of electric field at d' to that at d?

Accepted Answer

The electric potential V due to a point charge is inversely proportional to the distance (V ∝ 1/r). If the potential at d' is 2V, then d' = d/2. The electric field E is inversely proportional to the square of the distance (E ∝ 1/r^2). Therefore, the ratio of electric fields E(d')/E(d) = (d/d')^2 = (d/(d/2))^2 = 4.

Question 7

The electric field in a given location points along the -x-axis. This means that

Accepted Answer

The electric field points in the direction of decreasing potential.

Question 8

One member of a pair of parallel conducting plates has potential 17 V. An electron is ejected from that plate and travels the other plate, attaining a velocity of 5.0 × 10⁶ m/s in the process. What is the potential of the second plate? The mass of the electron is 9.11 × 10⁻³¹ kg.

Accepted Answer

The answer of One member of a pair of parallel...

Question 9

A pair of conductors are connected with a wire. Which of the following statements is necessarily true?

Accepted Answer

When two conductors are connected with a wire, they become equipotential surfaces since the charge will redistribute until they are at the same electric potential. Choices B and C are not necessarily true, as the electric field at the surface of each conductor depends on the amount of charge they carry, which could be different, and the conductors may have unequal charges. Choice D is also not necessarily true, as the conductors and wire could have a net charge, but their potentials would be the same.

Question 10

A parallel plate capacitor with circular plates having radius 0.07 m is filled with a dielectric of constant 2.2. If it is desired to store 12 × 10⁻⁶ J in this capacitor, what must the separation distance be for a voltage of 1.5 kV?

Accepted Answer

The energy stored in a capacitor is given by:$$U = \frac{1}{2}CV^2$$where:* U is the energy stored in joules (J)* C is the capacitance in farads (F)* V is the voltage across the capacitor in volts (V)The capacitance of a parallel plate capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$where:* C is the capacitance in farads (F)* $\epsilon_0$ is the permittivity of free space (8.85 × 10^-12 F/m)* A is the area of the plates in square meters (m^2)* d is the distance between the plates in meters (m)We are given the following information:* U = 12 × 10^-6 J* V = 1.5 kV = 1500 V* $\epsilon_r$ = 2.2* r = 0.07 mWe can first calculate the area of the plates:$$A = \pi r^2 = \pi (0.07 m)^2 = 0.0154 m^2$$We can then calculate the capacitance:$$C = \frac{\epsilon_0 \epsilon_r A}{d} = \frac{(8.85 	imes 10^{-12} F/m)(2.2)(0.0154 m^2)}{d}$$We can then substitute the given values into the equation for energy stored:$$12 	imes 10^{-6} J = \frac{1}{2}\left(\frac{(8.85 	imes 10^{-12} F/m)(2.2)(0.0154 m^2)}{d}ight)(1500 V)^2$$Solving for d, we get:$$d = 2.8 	imes 10^{-2} m = 2.8 cm$$Therefore, the correct answer is A.

Question 11

At a given location in space, the potential is known to be greater when one moves in any direction. To be more specific, no matter which direction one goes, a small step in that direction results in an increase in the potential. This means that the electric field at this location is

Accepted Answer

If the potential increases in every direction from a point, that point is at a local minimum of potential. The electric field, which points in the direction of greatest decrease of potential, is zero at a local minimum because there is no direction of decrease.

Question 12

Which of the following statements is true?

Accepted Answer

The answer of Which of the following statements is true?
...

Question 13

What is the force on a proton sitting midway between a pair of parallel conducting plates that are at potentials of −17 V and 55 V, respectively? Their separation distance is 1.5 cm.

Accepted Answer

The electric field between the plates is:$$E = \frac{V}{d} = \frac{55 V - (-17 V)}{1.5 cm} = 4833 V/cm$$The force on the proton is:$$F = qE = (1.6 	imes 10^{-19} C)(4833 V/cm) = 7.7 	imes 10^{-16} N$$

Question 14

A pair of parallel conducting plates have potentials of −17 V and 55 V, respectively. How much work needs to be done on an electron in order to move it from the 55 V to the −17 V plate?

Accepted Answer

The answer of A pair of parallel conducting plates have...

Question 15

The magnitude of the force on a proton sitting midway between a pair of parallel conducting plates that are at potentials of −17 V and 55 V, respectively, is 7.69 × 10⁻¹⁶ N. What is the separation distance between the plates?

Accepted Answer

The answer of The magnitude of the force on a...

Question 16

A parallel plate capacitor has air between the plates, which are circular with radius 0.07 m. The separation gap is 0.375 mm. What is the maximum energy that can be stored in this capacitor if air-breakdown is to be avoided? The breakdown threshold field is approximately 3 × 10⁶ N/C.

Accepted Answer

The maximum energy stored in a capacitor without causing air breakdown can be calculated using the formula for the energy stored in a capacitor, $U = \frac{1}{2}CV^2$, where $C$ is the capacitance and $V$ is the voltage. The breakdown voltage $V_{bd}$ can be found from the breakdown field $E_{bd}$ and the separation $d$, using $V_{bd} = E_{bd} \times d$. The capacitance $C$ of a parallel plate capacitor is given by $C = \epsilon_0 \frac{A}{d}$, where $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $A$ is the area of the plates, and $d$ is the separation between the plates. The area $A$ of a circular plate is $\pi r^2$, where $r$ is the radius of the plate.Given:- Radius $r = 0.07 \, \text{m}$- Separation $d = 0.375 \times 10^{-3} \, \text{m}$ (converted from mm to m)- Breakdown threshold field $E_{bd} = 3 \times 10^6 \, \text{N/C}$First, calculate the breakdown voltage $V_{bd}$:$$V_{bd} = E_{bd} \times d = 3 \times 10^6 \times 0.375 \times 10^{-3} = 1125 \, \text{V}$$Next, calculate the area $A$:$$A = \pi r^2 = \pi \times (0.07)^2 \approx 0.0154 \, \text{m}^2$$Then, calculate the capacitance $C$:$$C = \epsilon_0 \frac{A}{d} = 8.85 \times 10^{-12} \times \frac{0.0154}{0.375 \times 10^{-3}} \approx 3.65 \times 10^{-10} \, \text{F}$$Finally, calculate the maximum energy $U$:$$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 3.65 \times 10^{-10} \times (1125)^2 \approx 2.3 \times 10^{-4} \, \text{J}$$Therefore, the maximum energy that can be stored in the capacitor without causing air breakdown is approximately $2.3 \times 10^{-4} \, \text{J}$, which corresponds to choice B.

Question 17

An electron is projected horizontally with a speed of 5.0 × 10⁶ m exactly in the middle of two parallel plates at potentials of −40 V and 104 V, respectively. The plate separation is 3.0 cm. The plates are long enough that the electron, rather than passing completely between the plates, eventually strikes the plate sitting at 104 V potential. What is the speed of the electron when it strikes the plate? The mass of the electron is 9.11 × 10⁻³¹ kg.

Accepted Answer

The answer of An electron is projected horizontally with a...

Question 18

A pair of parallel conducting plates have potentials of −17 V and 55 V, respectively. An electron is ejected from the plate at potential −17 V and travels to the plate at 55 V. What is the change in the electron's potential energy as it makes this journey?

Accepted Answer

The change in potential energy (ΔU) is given by ΔU = qΔV, where q is the charge of the electron (-1.6 × 10^-19 C) and ΔV is the change in potential (55 V - (-17 V) = 72 V). Thus, ΔU = (-1.6 × 10^-19 C) * (72 V) = -1.152 × 10^-17 J, which rounds to -1.2 × 10^-17 J.

Question 19

An air-filled, parallel plate capacitor with circular plates of radius 0.07 m is used to store energy. If it is desired to store 12 × 10⁻⁶ J in this capacitor, what must the separation distance be for a voltage of 1.5 kV?

Accepted Answer

The answer of An air-filled, parallel plate capacitor with circular...

Question 20

The electric field in the dielectric-filled region between the plates of a capacitor is E. What is the charge density on the plates?

Accepted Answer

The charge density ($\sigma$) on the plates of a capacitor filled with a dielectric material is given by $\sigma = \epsilon_0 E$, where $\epsilon_0$ is the permittivity of free space and $E$ is the electric field. This matches choice D, which in plain text reads $\epsilon_0 E$, but due to a formatting error, it's listed as choice E here.

Quiz 17: Electric Potential