necessaryh
2021-03-23
Answered

Three charges are arranged as shown in Figure. Find the magnitude and direction of the electrostatic force on the charge at the origin.

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Mayme

Answered 2021-03-25
Author has **103** answers

Force between two charges on x-axis is repulsion force, so the force on charge at origin acts in negative x-direction

$F}_{1}=k(6\cdot {10}^{-9})\frac{5\cdot {10}^{-9}}{{\left(0.03\right)}^{2}}=8.98\cdot {10}^{9}\cdot (6\cdot {10}^{-9})\frac{5\cdot {10}^{-9}}{{\left(0.03\right)}^{2}$

$=2993.33\cdot {10}^{-9}$ N

Force between two charges on y-axis is attractive force, so the force on charge at origin acts in negative y-direction

$F}_{2}=k(3\cdot {10}^{-9})\frac{5\cdot {10}^{-9}}{{\left(0.01\right)}^{2}}=8.98\cdot {10}^{9}\cdot (3\cdot {10}^{-9})\frac{5\cdot {10}^{-9}}{{\left(0.01\right)}^{2}$

angle made by the net force with positive x-axis = 180 + taninverse$\frac{13470\cdot {10}^{-9}}{2993.33\cdot {10}^{-9}}=180+77.47=257.77$ degrees

magnitude of force$=\sqrt{{13470}^{2}+{2993.33}^{2}}\cdot {10}^{-9}=13798.5\cdot {10}^{-9}$ N

Force between two charges on y-axis is attractive force, so the force on charge at origin acts in negative y-direction

angle made by the net force with positive x-axis = 180 + taninverse

magnitude of force

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