Problem: Pictured on the right are three-point charges Q1 = 21.4 μC, Q2 = -30.6 μC, and Q3 = 57.3 μC arranged according to the figure on the right. A fourth point charge is located at point A with a charge of QA = 13.5 μC. Calculate the magnitude of the net force on the charge at point A. The answer should be in N.

FREE Expert Solution

Coulomb's law:

$\boxed{\mathbf{F}\mathbf{=}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$, where k is the coulomb's constant and r is the separation between the two charges. 

The magnitude of the net force:

$\boxed{{\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\sqrt{{{\mathbf{F}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{F}}_{\mathbf{y}}}^{\mathbf{2}}}}$

k = 9 × 10⁹ N·m²/C²

The force due to charge Q₁ is:

$\begin{array}{rcl}{\mathbf{F}}_{\mathbf{1}\mathbf{A}}& \mathbf{=}& \frac{\mathbf{k}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{A}}}{{{\mathbf{r}}_{\mathbf{1}\mathbf{A}}}^{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{(}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\mathbf{(}\mathbf{21}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}\mathbf{(}\mathbf{13}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}}{{\mathbf{(}\mathbf{80}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}\end{array}$

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Problem Details

Pictured on the right are three-point charges Q₁ = 21.4 μC, Q₂ = -30.6 μC, and Q₃ = 57.3 μC arranged according to the figure on the right. A fourth point charge is located at point A with a charge of Q_A = 13.5 μC. Calculate the magnitude of the net force on the charge at point A. The answer should be in N.