In this unit, you were learning about different yet similar laws of physics - Newton’s universal law of gravitation and Coulomb’s law of electrostatics. Describe one similarity and one difference for these two laws.
One similarity between Newton's law of gravitation and Coulomb's law of electrostatics is that both laws describe forces between objects based on their masses or charges, respectively.
However, a key difference between these laws is the nature of the force. In Newton's law of gravitation, the force is always attractive and depends on the product of the masses of the objects involved. On the other hand, in Coulomb's law of electrostatics, the force can be both attractive and repulsive and depends on the product of the charges of the objects involved.
Newton’s universal law of gravitation and Coulomb’s law of electrostatics. Describe one similarity and one difference for these two laws
One similarity between Newton's law of gravitation and Coulomb's law of electrostatics is that both laws describe inverse square relationships. In both laws, the force between two objects decreases with the square of the distance between them.
However, a key difference between these laws is the nature of the force and the properties of the objects involved. In Newton's law of gravitation, the force is always attractive and depends on the masses of the objects. On the other hand, in Coulomb's law of electrostatics, the force can be both attractive and repulsive and depends on the charges of the objects. Additionally, while gravity is always attractive and acts on all objects with mass, electrostatic forces only exist between objects with electric charge.
Refer to the following two images for the questions that follow.
A source charge with electric field lines radiating away from the source Two oppositely charged spheres are side by side with electric field lines surrounding both charges
Refer to Figure 1:
State whether the source charge is positively or negatively charged.
Figure 1 contains arrows that are outwardly pointing. By convention, state how the directions of field lines are detected.
Based on the description provided, the source charge in Figure 1 is positively charged.
In Figure 1, the arrows of the electric field lines are outwardly pointing. According to convention, this indicates that the electric field lines represent the direction a positive test charge would move if placed in the field.
Comparing Figure 1 and Figure 2, you can see Figure 2 does not have any arrowheads. Observe the sign of charges for each source charge, (in Figure 2) and state the direction the arrowheads should be pointing to make it accurate.
In Figure 2, the absence of arrowheads indicates that it is a diagram showing the electric field lines surrounding two oppositely charged spheres.
If the charges on the spheres are opposite as mentioned, the arrowheads on the electric field lines should point towards the positively charged sphere and away from the negatively charged sphere, in order for the diagram to accurately represent the electric field.
Examine the charge distribution in the diagram below. Illustration for the distribution of charges. Charge 1,2 and3 forma right triangle. Charge 1 and charge 2 form the base separated by 2.0 m. Charge 3 is directly above charge 1 separated by 2.0 m.
Determine the net force charge acting at q1 (+ 2.0 × 10-5
C), caused by q2 (-4.0 × 10-5
C) and q3 (-4.0 × 10-5
C). Be sure to include a free body diagram representing the force acting at q1 in your solution.
To determine the net force acting on q1, we need to calculate the individual forces between q1 and q2, as well as between q1 and q3, and then find the vector sum of these forces.
The force between two charges can be calculated using Coulomb's law:
F = (k * |q1| * |q2|) / r^2
Where F is the force between the charges, k is the Coulomb's constant (k = 9.0 x 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
Given the magnitudes of the charges and the distance between them, we can calculate the forces:
Force between q1 and q2:
|q1| = 2.0 x 10^-5 C
|q2| = 4.0 x 10^-5 C (negative sign indicates opposite charges)
r = 2.0 m
Using Coulomb's law:
F12 = (9.0 x 10^9 Nm^2/C^2 * |2.0 x 10^-5 C| * |4.0 x 10^-5 C|) / (2.0 m)^2
= -36.0 N
The negative sign indicates that the force between q1 and q2 is attractive.
Force between q1 and q3:
|q1| = 2.0 x 10^-5 C
|q3| = 4.0 x 10^-5 C (negative sign indicates opposite charges)
r = 2.0 m
Using Coulomb's law:
F13 = (9.0 x 10^9 Nm^2/C^2 * |2.0 x 10^-5 C| * |4.0 x 10^-5 C|) / (2.0 m)^2
= -36.0 N
The negative sign indicates that the force between q1 and q3 is attractive.
Now, we can find the net force acting on q1 by taking the vector sum of F12 and F13. Since both forces have the same direction (toward q1), we can simply add their magnitudes:
Net force = |F12| + |F13|
= |-36.0 N| + |-36.0 N|
= 72.0 N
Therefore, the net force acting on q1 is 72.0 N, directed toward q1 as shown in the free body diagram.
Determine the net electric field acting at q1.
To determine the net electric field acting at q1, we need to find the individual electric fields due to q2 and q3 and then combine them vectorially.
The electric field produced by a point charge can be calculated using the equation:
E = (k * |q|) / r^2
Where E is the electric field, k is the Coulomb's constant (k = 9.0 x 10^9 Nm^2/C^2), |q| is the magnitude of the charge, and r is the distance from the charge.
The electric field produced by q2 at q1 can be calculated as:
|q2| = 4.0 x 10^-5 C (negative sign indicates opposite charges)
r12 = 2.0 m
Using the equation for the electric field:
E2 = (9.0 x 10^9 Nm^2/C^2 * |4.0 x 10^-5 C|) / (2.0 m)^2
= 9.0 x 10^6 N/C
The electric field produced by q3 at q1 can be calculated as:
|q3| = 4.0 x 10^-5 C (negative sign indicates opposite charges)
r13 = 2.0 m
Using the equation for the electric field:
E3 = (9.0 x 10^9 Nm^2/C^2 * |4.0 x 10^-5 C|) / (2.0 m)^2
= 9.0 x 10^6 N/C
Now, we can find the net electric field at q1 by taking the vector sum of E2 and E3. Since both electric fields have the same direction (toward q1), we can simply add their magnitudes:
Net electric field = |E2| + |E3|
= |9.0 x 10^6 N/C| + |9.0 x 10^6 N/C|
= 18.0 x 10^6 N/C
Therefore, the net electric field acting at q1 is 18.0 x 10^6 N/C, directed toward q1.
An electron is in motion at 4.0 × 106
m/s horizontally when it enters a region of space between two parallel plates, as shown, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 102
N/C and separation between the charged plates is 2.0 cm.
Illustration of an electron passing through a uniform electric field. The field lines are pointing up
Referring to the above image, state the sign of the charge on each electric plate:
Sign of bottom plate - _______________________
Sign of the top plate - _______________________
Determine the acceleration of the electron between the plates.
Based on the description provided, the electron is entering the region of space starting at the negative plate and is deflected downwards. This implies that the bottom plate is negatively charged and the top plate is positively charged.
Therefore:
Sign of bottom plate: Negative (-)
Sign of top plate: Positive (+)
To determine the acceleration of the electron between the plates, we can use the equation for the force experienced by a charged particle in an electric field:
F = q * E
Where F is the force, q is the charge of the particle, and E is the electric field.
In this case, the charge of the electron is negative (-1.6 x 10^-19 C) and the electric field is provided as 4.0 x 10^2 N/C.
Using the equation, we can calculate the force experienced by the electron:
F = (-1.6 x 10^-19 C) * (4.0 x 10^2 N/C)
= -6.4 x 10^-17 N
Since force is equal to mass times acceleration (F = m * a), we can rearrange the equation to solve for the acceleration:
a = F / m
The mass of an electron is approximately 9.1 x 10^-31 kg.
a = (-6.4 x 10^-17 N) / (9.1 x 10^-31 kg)
= -7.03 x 10^13 m/s^2
Therefore, the acceleration of the electron between the plates is approximately -7.03 x 10^13 m/s^2, directed downwards as shown in the image.
Determine the horizontal distance travelled by the electron when it hits the plate.
To determine the horizontal distance traveled by the electron when it hits the plate, we can use the equations of motion to find the time it takes for the electron to hit the plate and then calculate the horizontal distance.
First, let's find the time it takes for the electron to hit the plate. We can use the equation of motion:
h = (1/2) * a * t^2
Where h is the vertical distance traveled by the electron (in this case, the distance between the plates), a is the acceleration, and t is the time.
Since the electron starts with a vertical velocity of 0 m/s and travels a distance of 2.0 cm (0.02 m) between the plates, we can rearrange the equation to solve for t:
0.02 m = (1/2) * (-7.03 x 10^13 m/s^2) * t^2
Simplifying the equation:
t^2 = -(0.02 m * 2) / (7.03 x 10^13 m/s^2)
t^2 = -2.85 x 10^-16 s^2
Since time cannot be negative, we will consider the positive square root:
t = √(2.85 x 10^-16 s^2)
t ≈ 1.69 x 10^-8 s
Now, using this time, we can calculate the horizontal distance traveled by the electron. Since the electron is moving horizontally with a velocity of 4.0 x 10^6 m/s:
distance = velocity * time
distance = (4.0 x 10^6 m/s) * (1.69 x 10^-8 s)
distance ≈ 6.76 x 10^-2 m
Therefore, the horizontal distance traveled by the electron when it hits the plate is approximately 6.76 x 10^-2 meters.
Determine the velocity of the electron as it strikes the plate
To determine the velocity of the electron as it strikes the plate, we need to know the time it takes for the electron to travel between the plates and the acceleration experienced by the electron.
We have already determined the time it takes for the electron to hit the plate, which is approximately 1.69 x 10^-8 s.
Now, we can use the equation of motion to calculate the final velocity of the electron:
v = u + a * t
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case:
Initial velocity (u): 4.0 x 10^6 m/s (given)
Acceleration (a): -7.03 x 10^13 m/s^2 (calculated previously)
Time (t): 1.69 x 10^-8 s (calculated previously)
Substituting the values into the equation:
v = (4.0 x 10^6 m/s) + (-7.03 x 10^13 m/s^2) * (1.69 x 10^-8 s)
Calculating the velocity:
v ≈ 4 x 10^6 m/s + (-1.19 x 10^6 m/s)
v ≈ 2.81 x 10^6 m/s
Thus, the velocity of the electron as it strikes the plate is approximately 2.81 x 10^6 meters per second.