Physics: 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?

1 1/5 48/40 = 1 8/40 1 8/40 Simplify to 1 1/5 Hope this helps!...

Three identical point charges of 2.0 μc are placed

maevemboucher78

09.03.2022

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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