Field Theory Worksheet

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Table of Contents


Quantities in field theory:

F = G \frac{m_1 \ m_2}{r^2}

\ F = Force of attraction between masses in N
\ G = Universal Gravitation constant
(\ G = 6.67x10-11 Nm2kg-2)
\ m_1 and \ m_2 = The two masses. (kg)
\ r = the distance separating their centers (m)

F = k \frac{q_1 \ q_2}{r^2}


\ F = Force of attraction or repulsion between charges (N)
\ k = "Coulomb" constant
(\ k = 8.99x109 Nm2C-2)
\ q_1 and \ q_2 = The two charges. (C)
\ r = the distance separating their centers (m)

g = \frac{F}{m}


\ g = Gravitational field strength (in N/kg)
\ F = Force exerted by the field on the mass m (N)
\ m = The mass that experiences the force (kg)

If you apply this formula to the one above, you can get this formula for the gravitational field due to a point or spherical mass:

g = G \frac{m}{r^2}

\ g = Gravitational field at distance \ r from point mass \ m.
\ g is always toward the mass

It is sad that this does not appear in the data packet.

E = \frac{F}{q}


\ E = Electrical field strength (in N/C)
\ F = Force exerted by the field on the charge q (N)
\ q = The charge that experiences the force (C)

If you apply this formula to the one above, you can get this formula for the electrical field due to a point or spherical charge:

E = k \frac{q}{r^2}

\ E = Electrical field at distance \ r from point charge \ q.
\ E is away from positive charge, but toward negative

It is sad that this does not appear in the data packet.

\Delta V = \frac{\Delta E_p}{m}

\ \Delta V = Change in gravitational potential in J/kg (?)
\ \Delta E_p = Change in gravitational potential energy (J)
\ m = mass that this applies to (kg)

\Delta V = \frac{\Delta E_p}{q}

\ \Delta V = Change in Electrical potential in J/C (Volts)
\ \Delta E_p = Change in electrical potential energy (J)
\ q = Charge that this applies to (kg)

V = - \frac{Gm}{r}


\ V = Gravitational potential (in J/kg) at a point due to a point mass.
\ G = Universal Gravitation constant
(\ G = 6.67x10-11 Nm2kg-2)
\ m = The mass creating the potential (kg)
\ r = The distance in meters (m) from the mass

Note that gravitational potentials are always negative. They represent the energy per unit mass needed to move a mass to infinitely far away (where the potential is zero)

V = \frac{kq}{r} = \frac{q}{4 \pi \varepsilon \,_o r}


\ V = Electrical potential (in J/C or Volts) at a point due to a point charge.
\ E = "Coulomb" constant
(\ k = 8.99x109 Nm2C-2)
\ q = The charge creating the potential (kg)
\ r = The distance in meters (m) from the charge
\varepsilon \,_o = The real constant that grownups use
(\varepsilon \,_o = 8.85x10-12 Nm-2C2)

You can show that

\frac{1}{4 \pi \varepsilon \,_o} = k

But you would need a calculator.

Note that electrical potentials are positive near a positive charge, and negative relative to a negative charge

g = - \frac{\Delta V}{\Delta r}


\ g = Gravitational field strength (N/kg)
\ \Delta V = Change in gravitational potential (J/kg)
\ \Delta r = Distance over which the gravitational potential changes (m)

E = - \frac{\Delta V}{\Delta x}


\ E = Electrical field strength (N/C)
\ \Delta V = Change in electrical potential (J/C) or Volts
\ \Delta x = Distance over which the electrical potential changes (m)

Problems

1. What is the gravitational field strength at the surface of a planet if 5.3 kg of mass experiences 67.2 N of downward force due to gravity?

The given values are 5.6 kg of mass (m) and a force (F) of 67.2 N. We are looking for gravitation field strength (g).

g = \frac{F}{m}

g = (67.2 N) / (5.6 kg)
g = 12.7 N/kg

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2. What force does a 34 N/kg gravitational field exert on a 72.5 kg mass?

The given values are a gravitational field (g) of 34 N/kg and a mass (m) of 72.5 kg. We are looking for force (F).

g = \frac{F}{m}

34 N/kg = F / 72.5 kg
F = 2465 N

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3. What is the force of gravity between a 674 kg object that is 7.15x106 m from earth’s center? (Earth has a mass of 5.97x1024 kg)

The given values are a distance (r) of 7.15E6 and two masses (m1, m2) of 674 kg and the Earth, which is 5.97E24 kg. We are looking for force (F).

F = G \frac{m_1 \ m_2}{r^2}

F = (6.67x10-11 Nm2kg-2)*(674 kg)*(5.97E24 kg) / (7.15E6 m)2
F = 5250 N

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4. At what distance from the center of the moon is the force of gravity on a 100.0 kg mass 5.00 N? (The moon has a mass of 7.35x1022 kg)

The given values are two masses (m1, m2) of 100 kg and the moon, which is 7.35E22 kg and a force (F) of 5 N. Since we are looking for distance (r), we will use the same formula as the last problem.

F = G \frac{m_1 \ m_2}{r^2}

5 N = (6.67E-11 Nm2kg-2)*(100 kg)*(7.35E22) / (r)2
r = 9901970 m
r = 9.90E6 m

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5. What is the change in gravitational potential (in J/kg) if it takes 812 J of energy to lift a 34 kg object from the floor to a shelf? If g is 9.81 N/kg, what is the shelf height?

The given values for the first part of the question are a mass (m) of 34 kg and 812 J of energy (\Delta E_p). We are looking for change in gravitational potential (\Delta V).

\Delta V = \frac{\Delta E_p}{m}

\Delta V = 812 J / 34 kg
\Delta V = 24 J/kg

The next part of the question asks for the height (h) and giving us Earth's gravity (g) of 9.81 N/kg. In order to solve for height (h), we will use the energy (\Delta E_p), the gravity (g) and the mass (m) in the equation

\Delta E_p = (m)(g)(h)
812 J = (34 kg)*(9.81 N/kg)*(h)
h = 2.43 m

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6. A 117 kg mass falls freely from rest through a gravitational potential difference of 45 J/kg. What potential energy in J does it turn into kinetic energy, and what is its final velocity?

The given values are a mass (m) of 117 kg and a change in gravitational potential (\Delta V) of 45 J/kg. The first part of the question asks to find potential energy (\Delta E_p).

\Delta V = \frac{\Delta E_p}{m}

45 J/kg = \Delta E_p / 117 kg
\Delta E_p = 5265 J = 5.3E3 J

Now that we have energy, we can use the formula that energy = 1/2mv2 to find the velocity, which the next part of the question asks for.

5265 J = .5*(117 kg)*(v2)
v = 9.487 m/s = 9.5 m/s

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7. What is the gravitational potential (in J/kg) 12 cm from the center of a 212 kg ball of lead?

The given values are .12 m of distance (r) and a mass (m) of 212 kg. We are looking for graviational potential (V).

V = - \frac{Gm}{r}

V = -(6.67E-11 Nm2kg-2)*(212 kg) / (.12 m)
V = -1.178E-7 J/kg = -1.2E-7 J/kg

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8. What is the gravitational potential (in J/kg) at the surface of the moon? (The moon has a mass of 7.35x1022 kg and a radius of 1.737x106 m)

The given values are the moon's mass (m) of 7.35E22 kg and the moon's radius (r) of 1.737E6. We are again looking for gravitational potential (V).

V = - \frac{Gm}{r}

V = -(6.67E-11 Nm2kg-2)*(7.35E22 kg) / (1.737E6)
V = -2822366 J/kg = -2.82E6 J/kg

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9. The earth has a mass of 5.97x1024 kg, and a radius of 6.38x106 m. What work does it take to bring a 5.00 kg object from the surface of the earth to an elevation of 4.15x106 m above the earth’s surface? How does that compare to the same calculation using \Delta E_p = mg\Delta h, where we assume g is a uniform 9.81 N/kg

The given values are a mass (m) of 5.97E24 kg, a radius (r) of 6.38E6 m and another radius (r) of 4.15E6 m above that. We also have another mass (m) of 5 kg. This is the same kind of thing as a moved charge problem, considering the 5 kg object is moving and the other object (Earth) isn't. We are looking for work (\Delta E_p), so first we will have to find change in gravitational potential (\Delta V).

V = - \frac{Gm}{r}

V1 = -(6.67E-11 Nm2kg-2)*(5.97E24 kg) / (6.38E6 m)
V1 = -62413636 J/kg
V2 = -(6.67E-11 Nm2kg-2)*(5.97E24 kg) / (6.38E6 + 4.15E6 m)
V2 = -37815669 J/kg
V2 - V1 = 24597967 J/kg = 2.46E7 J/kg

With our newfound change in gravitational potential (\Delta V), we can now solve for work (\Delta E_p).

\Delta V = \frac{\Delta E_p}{m}

2.46E7 J/kg = (\Delta E_p) / (5 kg)
(\Delta E_p) = 1.23E8 J

The next part of the question asks us the find the same thing (\Delta E_p), but this time by using the formula (\Delta E_p) = mgh

(\Delta E_p) = (5 kg)*(9.81 m/s2)*(4.15E6 m)
(\Delta E_p) = 2.04E8 J

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10. An object approaches the moon. ( Moon’s mass: 7.35 x 1022 kg, radius: 1.737 x 106 m) If the object is going 1210 m/s when it is 2.26 x 106 m from the moon’s center, what is its speed when it strikes the surface?

The given values are we have the moon's mass (m) of 7.35E22, the moon's radius (r) of 1.737E6, an object's velocity (v) of 1210 m/s and that object's distance (r) from the moon's center of 2.26E6. The question is looking for another velocity (v2).

The object has an original velocity that we have and a final velocity that we need to solve for. The equation with velocity in it for this unit is 1/2mv2, so we know we're going to have to deal with two of those equations, one for the initial velocity and one for the final velocity. Both velocities differ by a factor of potential energy (\Delta E_p). The energy can be found by the formula

\Delta V = \frac{\Delta E_p}{m}

In turn, that means \Delta E_p = {\Delta V}{m}. Since the first velocity has that potential energy (the second one doesn't because it's already hit the surface, items on the surface don't have any potential energy), we can use 1/2mv12 + {\Delta V}{m} = 1/2mv12 to find the second velocity. First, however, we need to find our \Delta V.

V = - \frac{Gm}{r}

V1 = -(6.67E-11 Nm2kg-2)*(7.35E22 kg) / (1.737E6 m)
V1 = -2822366 J/kg
V2 = -(6.67E-11 Nm2kg-2)*(7.35E22 kg) / (2.26E6 m)
V2 = -2169225 J/kg
V2 - V1 = 653140 J/kg

With our newfound change in gravitational potential of 653140 J/kg, we can now solve for our other velocity using 1/2mv12 + {\Delta V}{m} = 1/2mv22. However, the object we are given doesn't have a mass. If we look at the equation we are going to use to solve for the velocity, the mass cancels out on both sides. Divide everything by (m) and the (m) goes away completely, leaving us with

1/2v12 + {\Delta V} = 1/2v22
1/2(1210 m/s)2 + (653140 J/kg) = 1/2v2
v2 = 1664 m/s = 1.66E3 m/s

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11. How much work does it take to move two 4.5x106 kg spheres whose centers are separated by 3.5 m initially, so that their centers are separated by 7.8 m?

The given values are two masses (m) of 4.5E6 kg, the original distance (r) of 3.5 and the moved distance of 7.8m (r). This is the same thing as a moved charge problem, but in terms of mass and gravity. We are looking for work (\Delta E_p) so we will first find the gravitational potential (\Delta V) for each (the change in that is what is required to solve for our work).

V = - \frac{Gm}{r}

V1 = -(6.67x10-11 Nm2kg-2)*(4.5E6 kg) / (3.5 m)
V1 = -8.58E-5 J/kg
V2 = -(6.67x10-11 Nm2kg-2)*(4.5E6 kg) / (7.8 m)
V2 = -3.85E-5 J/kg
V2 - V1 = 4.73E-5 J/kg

We can use our new value, 4.73E-5 J/kg of the change of gravitational potential (\Delta V) to find our work (\Delta E_p).

\Delta V = \frac{\Delta E_p}{m}

4.73E-5 J/kg = \Delta E_p / (4.5E6 kg)
\Delta E_p = 212 J

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12. What is the gravitational field strength on the surface of the moon with a radius of 1.737x106 m, and a mass of 7.35x1022 kg?

We are going to use an inferential formula here. The data packet gives us these two formulas:

F = G \frac{m_1 \ m_2}{r^2}

and

g = \frac{F}{m}

So we can infer that

g = G \frac{m}{r^2}

This is not in the data packet, so you need to memorize it, or be able to infer it

So we are all set here:

g = 6.67\times10^{-11} \frac{7.35\times10^{22}}{(1.737\times10^6)^2} = 1.624850978 = 1.62 N/kg

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13. What is the radius of a neutron star with a gravitational field strength of 3.4x1013 N/kg, and a mass of 8.13x1031 kg?

The given values are a gravitational field strength (g) of 3.4E13 and a mass (m) of 8.13E31 kg. Since we are looking for radius (r) we can find it by using

g = G \frac{m}{r^2}

3.4E13 N/kg = (6.67E-11 Nm2kg-2)*(8.13E31 kg) / (r)2
r = 12629 m = 13 km

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14. The top of a hill is 420 J/kg higher in gravitational potential than the bottom. What work does it take to lift a 5.0 kg object from the bottom of the hill to the top? If the g is 9.81 N/kg, how high is the hill?

For the first part of the question, the given values are a gravitational potential difference (\Delta V) of 420 J/kg and a mass (m) of 5 kg. We are looking for work (\Delta E_p).

\Delta V = \frac{\Delta E_p}{m}

420 J/kg = \Delta E_p / 5 kg
\Delta E_p = 2100 J

With our new value of 2100 J of work, we can use that and our gravity (g) of 9.8 to find our height using \Delta E_p = mgh

2100 J = (5 kg)*(9.8 m/s2)*(h)
h = 42.857 m = 43 m

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15. On the planet Zot, a hill 23.1 m high represents a potential change of 416 J/kg of gravitational potential. What is the g on planet Zot?

The given values are a change in gravitational potential (\Delta V) of 416 J/kg and a distance (r) of 23.1 m. We are looking for gravity (g). The formula we have to use is therefore

g = - \frac{\Delta V}{\Delta r}

Since we need a change in distance and the hill is 23.1 m up, an object falling would go from 23.1 m high to 0 m high. Since we use final - initial to find the change in something, 0 m - 23 m = -23 m. -23 m is our \Delta r.

g = -(416 J/kg) / (-23.1 m)
g = 18 N/kg

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16. What is the electrical field strength near a charged object if 2.3 μC of charge experiences .018 N of upward force due to the electric field?

The given values are 2.3E-6 C of charge (q) and .018 N of force (F). We are looking for electrical field (E). The formula we can use to find this is

E = \frac{F}{q}

By dividing .018 by 2.3E-6, we get 7826 N/C (electrical field). Because we need two significant figures, the answer is simplified to 7.8E3.

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17. What force does a 5200 N/C electrical field exert on a proton?

The given values are 5200 N/C of electrical field (E) and since we are given that the item is a proton, we can find the charge in our packet, 1.60E-19 (elementary charge). Elementary charge is the same for both protons and electrons. Since we are looking for force (F) and we have E and q, we will use the same formula from the last problem.

E = \frac{F}{q}

5200 N = F / (1.60E-19 C)

Solve for F by multiplying both sides and we get the answer 8.3E-16 N/C.

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18. A 34.8 μC charge is 45.0 cm to the right of a -12.7 μC charge. What is the force on the leftmost charge?

The given values are the two charges, 34.8E-6 C (q1) and -12.7E-6 (q2) and .45m of distance (r). Since we are looking for force (F) and have (q1), (q2) and (r), we can use the formula:

F = k \frac{q_1 \ q_2}{r^2}

Given k is 8.99E9

F = (8.99E9 Nm2C-2)*(34.8E-6 C)*(12.7E-6 C) / (.45m)2

We don't need to worry about whether or not the charges are positive or negative. The only thing that positive or negative charge changes is what direction the force is in, not the actual force itself.

F = 19.6 N

Since force is a vector, we do have to find what direction the force is. Since positive and negative charges attract, both charges are going to have a force towards each other. We are looking for the direction of the force on the leftmost charge. If they are attracted to each other, the charge on the left will be attracted to the right.

F = 19.6 N to the right.

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19. How far apart do two 1.00 C charges need to be so that they experience a force of repulsion of 1.00 N

The given values are two charges of 1 C (q1 q2) and a force of 1 N (F). Since we are looking for distance (r), we can use the formula

F = k \frac{q_1 \ q_2}{r^2}

Given k = 8.99E9

1 = 8.99E9 Nm2C-2* 1 C * 1 C / r2

Solving for r, we will come to

r2 = 8.99E9

r = 9.48E4 m

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20. What is the change in electrical potential if it takes 812 J of energy to move a 15 μC charge from point A to point B?

The given values are 812 J of energy (\Delta E) and a charge of 15E-6 (q). We are looking for electric potential in Volts (\Delta V). To solve this, we can use the formula

\Delta V = \frac{\Delta E}{q}

\Delta V = 812 J / 15E-6 C

\Delta V = 5.4E7 J/C

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21. A proton accelerates from rest through an electric potential of 24.0 V. What is its change in potential energy, and what is its final velocity?

The given values are electric potential in 24.0 V (\Delta V), the charge of a proton, 1.60E-19 C (q), and the rest mass of a proton, 1.673E-27 kg (m). The rest mass of a proton and elementary charge can be found in the data packet.

To solve the first part of this problem, we are looking for change in potential energy (\Delta E). We will use the same formula to find that as we did in the last problem.

\Delta V = \frac{\Delta E}{q}

24 V = \Delta E / 1.60E-19 C

\Delta E = 3.84E-18 J

The next part of the problem asks for velocity. Since we now have a newfound value, energy (potential converts to kinetic), we can use

E_k = \frac{1}{2}mv^2

Given the mass of a proton is 1.673E-27 kg,

3.84E-18 J = (.5)*(1.673E-27 C)*v2

v = 6.78E4 m/s

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22. What is the electrical potential 7.8 cm from the center of a 0.215 μC charge?

The given values are .078 m distance (r) and 0.215E-6 C (q). Since we are looking for electric potential (V), we can use

V = \frac{kq}{r}

Given that k = 8.99E9

V = (8.99E9 Nm2C-2)*(0.215E-6 C) / (.078 m)

V = 2.5E4 V

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23. A sphere has a 2.1 μC charge spread evenly over its surface. What is the sphere’s radius if the electric potential on the surface is 48,000 V?

The given values are 2.1E-6 C of charge (q) and 48,000 V of electric potential (V). Since we are looking for the distance of the radius (r), we can use the same formula from the last problem

V = \frac{kq}{r}

Given that k = 8.99E9

48,000 V = (8.99E9 Nm2C-2)*(2.1E-6 C) / r

r = .39 m

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24. A 25.1 μC charge is 67.0 cm to the right of a 16.8 μC charge. What work would it take to bring them closer so that they are separated by only 45.0 cm? If they each have a mass of 12.1 grams, what is their speed when they are very far away, if they are released from a distance of 45.0 cm? (Neglect other forces)

The given values are 25.1E-6 C of charge (q), 16.8E-6 of charge (q), and .67 m of distance originally (r). We are looking for work in J. Since work is the same as energy (both in J) we are solving for (\Delta E_p). The only electrical formula that we can use to find energy is

\Delta V = \frac{\Delta E}{q}

In order to find the energy, we need charge (q) and change in voltage (\Delta V). This is a moved charge problem, so we can make one of the charges moving the full distance from .67m to .45m and the other one stationary. If we decide to use the charge 25.1E-6 charge as the moved one, we first must find the change in voltage before we can find the energy. From the given charge and the given distance, we can find the voltage at .67m and .45m. In short, finding the difference of the two will give us the change in voltage. So we'll use this formula:

V = \frac{kq}{r}

V1 = (8.99E9 Nm2C-2)*(25.1E-6 C) / (.67m)
V1 = 3.37E5 V
V2 = (8.99E9 Nm2C-2)*(25.1E-6 C) / (.45m)
V2 = 5.01E5 V
V2-V1 = 1.64E5 V
\Delta V = 1.64E5 V

Using \Delta V = \frac{\Delta E}{q} with (q) being the stationary charge

1.64E5 V = (\Delta E_p) / (16.8E-6 C)

\Delta E_p = 2.76 J

For the next part of this problem, we are looking for velocity in m/s (v). In order to obtain velocity, we will first need energy to apply E = 1/2mv2. We will use the same formula as we did in the first part to obtain the energy, however, instead of using .67m and .45m, we will use .45m and infinity (really far away). A note to add, anything divided by infinity is 0.

V1 = (8.99E9 Nm2C-2)*(25.1E-6 C) / (infinity m)
V1 = 0 V
V2 = (8.99E9 Nm2C-2)*(25.1E-6 C) / (.45m)
V2 = 5.01E5 V
V2-V1 = 5.01E5 V
\Delta V = 5.01E5 V

Using \Delta V = \frac{\Delta E}{q} with (q) being the other charge

50.1E5 V = (\Delta E_p) / (16.8E-6 C)

\Delta E_p = 8.42 J

However, that's the total energy both of them have. Since they are both the same mass, they will both use the energy equally. We will split the energy in half. (8.42 J) / 2 = 4.21 J. Given either one of them has 4.66 J and a mass of .012 kg, we can now use E = 1/2mv2 to find the velocity.

4.21 J = (.5)*(.012 kg)*(v2)
v = 26.5 m/s


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25. Three 115 gram 25.0 μC charges occupy the corners of an equilateral triangle 30.0 cm on a side. If the charges are released simultaneously, what is their speed when they are very far away, assuming no other force acts on them? (Bring the charges in from infinity one by one, add the work together…)

The given values are the three charges (q) of 25E-6, their masses (m) of .115 kg and the distance they are from each other (r) of .3 m. We are looking for velocity (v). In order to find velocity, we need to use the formula \Delta E_p<math> = 1/2mv<sup>2</sup>. To find <math>\Delta E_p, we need to use

\Delta V = \frac{\Delta E_p}{q}

In order to use that, we need to find the change in electric potential \Delta V. So, for all three charges, we will use

V = \frac{kq}{r}

V = (8.99E9 Nm2C-2)*(25E-6 C) / (.3 m)
V = 739167 V

Since we need this for all three charges, we must multiply our voltage by 3. (3)*(739167 V) = 2247500 V. Now we can solve for the work (energy).

2247500 V = \Delta E_p / (25E-6 C)
\Delta E_p = 56.1875 J

Now, however, since that is the total work, we need to divide this answer by three in order to get the velocity of an individual charge. (56.1875 J) / (3) = 18.7 J. In theory, we never needed to multiply by 3 in the first place if we were going to divide by 3 afterwards, but if there were more than 3 charges this problem would differ. So learn it this way.

\Delta E_p = 1/2mv2
18.7 J = 1/2*(.115 kg)*(v)2
v = 18.048 m/s = 18.0 m/s

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26. How much work does it take to move two 4.5 μC spheres whose centers are separated by 3.5 m initially, so that their centers are separated by 7.8 m?

The given values are two charges (q) of 4.5E-6 C, separated by 3.5 m (r) and 7.8 m (r). This is another moved charge problem, this time looking for work (\Delta E_p). Given that we need to find change in voltage (\Delta V), we will first need to use the formula to find the voltage for both distances:

V = \frac{kq}{r}

V1 = (8.99E9 Nm2C-2)*(4.5E-6 C) / (3.5 m)
V1 = 11558.6 V
V2 = (8.99E9 Nm2C-2)*(4.5E-6 C) / (7.8 m)
V2 = 5186.5 V
V2 - V1 = -6372.1 V

With (\Delta V) and our charge (q), we can now find our work (\Delta E_p) with the formula:

\Delta V = \frac{\Delta E_p}{q}

-6372.1 V = (\Delta E_p)*(4.5E-6 C)
\Delta E_p = -.0287 J

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27. What is the electric field 3.5 cm to the right of a 0.0137 μC charge?

The given values are .035 m (r) and 0.0137E-6 C of charge (q). Since we are looking for the electric field (E) we will use the formula

E = k \frac{q}{r^2}

E = (8.99E9Nm2C-2)*(0.0137E-6 C) / (0.035 m)2
E = 100541 N/C

It is also necessary to express the direction in which the electric field is. Electric field goes by the rule WWPCD (what would positive charge do). The electric field is to the right of a positive charge. If the electric field was a positive charge (because it is a positive electric field), it would move away from the original charge, because two positive charges repel each other. Hence, it would move to the right.

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28. There is a 12,350 N/C upward electric field 15.6 cm below what charge?

The given values are an electric field (E) of 12,350 N/C and a distance of .156 m (r). Since we are looking for charge (q), we will use the same formula as last equation.

E = k \frac{q}{r^2}

12,350 N/C = (8.99E9 Nm2C-2)*(q) / (.156 m)2
q = 3.34E-8

The final step is to find if the charge is positive or negative. Since a positive, upward electric field is below the charge, the field is pointing towards the charge. If the field were a positive charge (WWPCD), the only way for it to be attracted to our (q) is for our (q) to be a negative charge, since opposites attract.

q = -3.34E-8

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29. Two parallel plated have a potential of 600. Volts across them. What work does it take to move 2.3 C of charge from one plate to the other? If the plates have an electric field intensity of 14,500 V/M between them, what is their separation?

The given values for the first part of the problem are 600. V of electric potential (V) and a charge (q) of 2.3 C. Since we are looking for work (\Delta E_p), we will use the equation

\Delta V = \frac{\Delta E_p}{q}

600 V = (\Delta E_p)*(2.3 C)
\Delta E_p = 1380 J

The next part of the problem asks for the separation between plates (\Delta x). Given the voltage (V) of 600 and the newfound given electric field (E) of 14,500 V/m, we can use the formula

E = - \frac{\Delta V}{\Delta x}

14,500 V/m = 600 V / (\Delta x)
\Delta x = .0414 m

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30. Two parallel plates are separated by 3.1 mm, and have 710. Volts of electric potential across them. What is the electrical field intensity between the plates?

The given values are the separation of the plates (\Delta x), .0031 m, and the voltage (V) of 710 V. Since we are looking for electric field (E), we can use the same formula as the second part of the last problem.

E = - \frac{\Delta V}{\Delta x}

E = (710 V) / (.0031 m)
E = 229032 V/m

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