Defined Constants The following constants can be used: ke = 8.99×109 Nm2/C2 qe =
ID: 1524737 • Letter: D
Question
Defined Constants
The following constants can be used:
ke = 8.99×109 Nm2/C2
qe = 1.6×10–19 C
mp = 1.67×10–27 kg
me = 9.11×10–31 kg
eps0 = 8.85×10–12 C2/(Nm2)
g = 9.80 m/s2
1.) Consider the picture displayed. The charge of the middle charge is 2.49 C. Determine the electric field strength at a point 1.00 cm to the left of the middle charge.
Additional input values q2 (C): 2.23, 2.61
Defined symbols:
q2 charge of the middle charge in C
q1 charge of the left chargein C
q3 charge of the right charge in C
d12 distance between q1 and q2 in m
d23 distance between q2 and q3 in m
The answer is E electric field strength at a point 1.00 cm to the left of the middle charge in N/C
Hints:
You will have to calculate the electric field of each of the three charges at the point 1.00 cm left of the middle charge and add the results together. You add the electric field just like you added the forces in problem 4.
It can be helpful to invent a small positively charged ball and place it at the point where you wish to find the field. The only thing that's different between finding the force on this ball and the electric field at this point is that the electric field doesn't include the charge of the ball.
Range of answers are within: –1.15×108 N/C to +2.00×107 N/C
2.) This is a continuation of the last problem. The input values are the same.
Consider the picture. The charge of the middle charge is 2.49 C. If a charge of +2.00 C is placed at a point 1.00 cm to the left of the middle charge, what are the magnitude and direction of the force on it? (Let a force to the right be positive and a force to the left be negative.)
Additional input values q2 (C): 2.23, 2.61
Defined symbols:
q charge of the test charge in C
q2 charge of the middle charge in C
q1 charge of the left charge in C
q3 charge of the right charge in C
d12 distance between q1 and q2 in m
d23 distance between q2 and q3 in m
The answer is F1 magnitude of the force on the +2.00 C charge in N
Hints:
Since we know the electric field from the previous problem (Be sure to use the E from the correct input!), the force is just F=qE. Because you know the electric field strength at that point, you can recalculate the force on any charge placed there without having to add all the vectors again.
Range of answers are within: –230 N to +40.0 N
3.) Consider the picture. The charge of the middle charge is 2.49 C. If a charge of –2.00 C is placed at a point 1.00 cm to the left of the middle charge, what are the magnitude and direction of the force on it? (Let a force to the right be positive and a force to the left be negative.)
Additional input values q2 (C): 2.23, 2.61
Defined symbols:
q charge of the test charge in C
q2 charge of the middle charge in C
q1 charge of the left chargein C
q3 charge of the right charge in C
d12 distance between q1 and q2 in m
d23 distance between q2 and q3 in m
The answer is F2 magnitude of the force on the –2.00 C charge in N
Hints:
Since we know the electric field from the previous problem, the force is just F=qE. Because you know the electric field strength at that point, you can recalculate the force on any charge placed there without having to add all the vectors again.
Range of answers are within: –40 N to +230.0 N
1.) Consider the picture displayed. The charge of the middle charge is 2.49 C. Determine the electric field strength at a point 1.00 cm to the left of the middle charge.
Additional input values q2 (C): 2.23, 2.61
Defined symbols:
q2 charge of the middle charge in C
q1 charge of the left chargein C
q3 charge of the right charge in C
d12 distance between q1 and q2 in m
d23 distance between q2 and q3 in m
The answer is E electric field strength at a point 1.00 cm to the left of the middle charge in N/C
Hints:
You will have to calculate the electric field of each of the three charges at the point 1.00 cm left of the middle charge and add the results together. You add the electric field just like you added the forces in problem 4.
It can be helpful to invent a small positively charged ball and place it at the point where you wish to find the field. The only thing that's different between finding the force on this ball and the electric field at this point is that the electric field doesn't include the charge of the ball.
Range of answers are within: –1.15×108 N/C to +2.00×107 N/C
2.) This is a continuation of the last problem. The input values are the same.
Consider the picture. The charge of the middle charge is 2.49 C. If a charge of +2.00 C is placed at a point 1.00 cm to the left of the middle charge, what are the magnitude and direction of the force on it? (Let a force to the right be positive and a force to the left be negative.)
Additional input values q2 (C): 2.23, 2.61
Defined symbols:
q charge of the test charge in C
q2 charge of the middle charge in C
q1 charge of the left charge in C
q3 charge of the right charge in C
d12 distance between q1 and q2 in m
d23 distance between q2 and q3 in m
The answer is F1 magnitude of the force on the +2.00 C charge in N
Hints:
Since we know the electric field from the previous problem (Be sure to use the E from the correct input!), the force is just F=qE. Because you know the electric field strength at that point, you can recalculate the force on any charge placed there without having to add all the vectors again.
Range of answers are within: –230 N to +40.0 N
3.) Consider the picture. The charge of the middle charge is 2.49 C. If a charge of –2.00 C is placed at a point 1.00 cm to the left of the middle charge, what are the magnitude and direction of the force on it? (Let a force to the right be positive and a force to the left be negative.)
Additional input values q2 (C): 2.23, 2.61
Defined symbols:
q charge of the test charge in C
q2 charge of the middle charge in C
q1 charge of the left chargein C
q3 charge of the right charge in C
d12 distance between q1 and q2 in m
d23 distance between q2 and q3 in m
The answer is F2 magnitude of the force on the –2.00 C charge in N
Hints:
Since we know the electric field from the previous problem, the force is just F=qE. Because you know the electric field strength at that point, you can recalculate the force on any charge placed there without having to add all the vectors again.
Range of answers are within: –40 N to +230.0 N
6.00 -2.00 AC 3.00 cm 2.00 cmExplanation / Answer
The field at the given point will be the sum of fields due to the individual charges.
E = E1 + E2 + E3
E = k q/r1^2 + k q2/r2^2 + k q3/r3^2
E = 9 x 10^9 [6 x 10^-6/0.02^2 + 2.49 x 10^-6/0.01^2 + (-2 x 10^-6)/0.03^2]
E = 9 x 10^3 [ 15000 + 24900 - 2222 ] = 3.39 x 10^8 N/C
Hence, E = 3.39 x 10^8 N/C
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