Three charges are placed along a line segment at points A, B and C (as in the figure). Charges are placed at points A and C. Let QA = 85 μµC and qc = 25 μC. Note that distance L₁ = 0.92 m and distance L₂ = 0.46 m. Recall that 1 μC = 10-6 C. What is the electric field at point B due to the charges at A and C? Enter a positive number if directed toward the right and a negative number is directed to the left. E = N/C If a charge qB = -140 μC is placed at point B, what is the net force acting on this charge due to charges at A and C? Indicate the direction of the force (positive = right, negative = left).

College Physics
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Chapter18: Electric Charge And Electric Field
Section: Chapter Questions
Problem 42PE: (a) Find the total electric field at x = 1.00 cm in Figure 18.52(b) given that q =5.00 nC. (b) Find...
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Three charges are placed along a line segment at points A, B
and C (as in the figure). Charges are placed at points A and C.
Let QA = 85 μC and qc = 25 μC. Note that distance L₁ = 0.92 m
and distance L₂ = 0.46 m. Recall that 1 μC = 10-6 C.
What is the electric field at point B due to the charges at A and
C? Enter a positive number if directed toward the right and a
negative number is directed to the left.
E =
If a charge qB = -140 μC is placed at point B, what is the net
force acting on this charge due to charges at A and C? Indicate
the direction of the force (positive right, negative = left).
F =
N/C
A
N
4₁
B
L₂
C
Transcribed Image Text:Three charges are placed along a line segment at points A, B and C (as in the figure). Charges are placed at points A and C. Let QA = 85 μC and qc = 25 μC. Note that distance L₁ = 0.92 m and distance L₂ = 0.46 m. Recall that 1 μC = 10-6 C. What is the electric field at point B due to the charges at A and C? Enter a positive number if directed toward the right and a negative number is directed to the left. E = If a charge qB = -140 μC is placed at point B, what is the net force acting on this charge due to charges at A and C? Indicate the direction of the force (positive right, negative = left). F = N/C A N 4₁ B L₂ C
Expert Solution
Step 1

Coulomb's law of electrostatics states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In simpler terms, the closer two charged objects are to each other, the stronger the force between them, and the larger the charges on the objects, the stronger the force between them.

The electric field at a point is the force per unit charge that would be experienced by a small positive test charge placed at that point. We can find the electric field at a point by using Coulomb's law, The mathematical representation of an electric field due to a single point charge +Q is given by,

E= 1(4*pi*ε₀) Q/r^2,

where E is the electric field vector at the point of interest, k=1(4*pi*ε₀) is Coulomb's constant (approximately 9 x 10^9 N·m^2/C^2), q is the magnitude of the point charge, r is the distance between the point charge and the point of interest, and u is a unit vector pointing in the direction of the electric field.

The electric field is a vector quantity, which means that it has both magnitude and direction. The magnitude of the electric field is proportional to the magnitude of the point charge and inversely proportional to the square of the distance between the charge and the point of interest. The direction of the electric field is radial and points away from a positive charge and towards a negative charge.

For a system of multiple point charges, the total electric field at a point in space can be found by vector addition of the electric fields due to each individual charge. The resulting electric field vector is the vector sum of all the individual electric fields at the point of interest.

 

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