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fgiga [73]
3 years ago
5

Three identical point charges of 2.0 μC are placed on the x-axis. The first charge is at the origin, the second to the right at

x = 50 cm, and the third is at the 100 cm mark. What are the magnitude and direction of the electrostatic force which acts on the charge at the origin?

Physics
1 answer:
sergij07 [2.7K]3 years ago
3 0

Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C

r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m

r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m

F₁ = magnitude of force by charge at B on charge at origin

F₂ = magnitude of force by charge at C on charge at origin

Magnitude of force by charge at B on charge at origin

F_{1}=\frac{kQ^{2}}{r_{1}^{2}}

inserting the values

F_{1}=\frac{(9\times 10^{9})(2 \times 10^{-6})^{2}}{0.5^{2}}

F₁ = 0.144 N


Magnitude of force by charge at C on charge at origin

F_{2}=\frac{kQ^{2}}{r_{2}^{2}}

inserting the values

F_{2}=\frac{(9\times 10^{9})(2 \times 10^{-6})^{2}}{1^{2}}

F₂ = 0.036 N

Net force on the charge at the origin is given as

F = F₁ + F₂

F = 0.144 + 0.036

F = 0.18 N

from the diagram , direction of net force is towards left or negative x-direction.


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