Let's first find the magnitude of the force with which the charge q 1 q_1 q 1 q 3 q_3 q 3
F 13 = k q 1 q 3 r 13 2 = k q 1 q 3 ( r 12 2 + r 23 2 ) 2 , F_{13}=\dfrac{kq_1q_3}{r_{13}^2}=\dfrac{kq_1q_3}{(\sqrt{r_{12}^2+r_{23}^2})^2}
, F 13 = r 13 2 k q 1 q 3 = ( r 12 2 + r 23 2 ) 2 k q 1 q 3 , F 13 = 9 ⋅ 1 0 9 N m 2 C 2 ⋅ 1.5 ⋅ 1 0 − 9 C ⋅ 5 ⋅ 1 0 − 9 C ( ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 ) 2 = 1.205 ⋅ 1 0 − 9 N . F_{13}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot1.5\cdot10^{-9}\ C\cdot5\cdot10^{-9}\ C}{(\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2})^2}=1.205\cdot10^{-9}\ N. F 13 = ( ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 ) 2 9 ⋅ 1 0 9 C 2 N m 2 ⋅ 1.5 ⋅ 1 0 − 9 C ⋅ 5 ⋅ 1 0 − 9 C = 1.205 ⋅ 1 0 − 9 N . Similarly, let's find the magnitude of the force with which the charge q 2 q_2 q 2 q 3 q_3 q 3
F 23 = k q 2 q 3 r 23 2 , F_{23}=\dfrac{kq_2q_3}{r_{23}^2}, F 23 = r 23 2 k q 2 q 3 , F 23 = 9 ⋅ 1 0 9 N m 2 C 2 ⋅ 3.2 ⋅ 1 0 − 9 C ⋅ 5 ⋅ 1 0 − 9 C ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 = 7.2 ⋅ 1 0 − 9 N . F_{23}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot3.2\cdot10^{-9}\ C\cdot5\cdot10^{-9}\ C}{(\sqrt{(2\ m)^2+(4\ m)^2})^2}=7.2\cdot10^{-9}\ N. F 23 = ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 9 ⋅ 1 0 9 C 2 N m 2 ⋅ 3.2 ⋅ 1 0 − 9 C ⋅ 5 ⋅ 1 0 − 9 C = 7.2 ⋅ 1 0 − 9 N . Let's find the projections of forces F 13 F_{13} F 13 F 23 F_{23} F 23 x x x y y y
F 13 , x = F 13 ⋅ r 23 r 13 = 1.205 ⋅ 1 0 − 9 N ⋅ ( 2 m ) 2 + ( 4 m ) 2 ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 , F_{13,x}=F_{13}\cdot\dfrac{r_{23}}{r_{13}}=1.205\cdot10^{-9}\ N\cdot\dfrac{\sqrt{(2\ m)^2+(4\ m)^2}}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}}, F 13 , x = F 13 ⋅ r 13 r 23 = 1.205 ⋅ 1 0 − 9 N ⋅ ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 ( 2 m ) 2 + ( 4 m ) 2 , F 13 , x = 7.2 ⋅ 1 0 − 10 N . F_{13,x}=7.2\cdot10^{-10}\ N. F 13 , x = 7.2 ⋅ 1 0 − 10 N . F 13 , y = − F 13 ⋅ r 12 r 13 = − 1.205 ⋅ 1 0 − 9 N ⋅ 6 m ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 , F_{13,y}=-F_{13}\cdot\dfrac{r_{12}}{r_{13}}=-1.205\cdot10^{-9}\ N\cdot\dfrac{6\ m}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}}, F 13 , y = − F 13 ⋅ r 13 r 12 = − 1.205 ⋅ 1 0 − 9 N ⋅ ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 6 m , F 13 , y = − 9.66 ⋅ 1 0 − 10 N . F_{13,y}=-9.66\cdot10^{-10}\ N. F 13 , y = − 9.66 ⋅ 1 0 − 10 N . F 23 , x = − F 23 ⋅ r 23 r 13 = − 7.2 ⋅ 1 0 − 9 N ⋅ ( 2 m ) 2 + ( 4 m ) 2 ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 , F_{23, x}=-F_{23}\cdot\dfrac{r_{23}}{r_{13}}=-7.2\cdot10^{-9}\ N\cdot\dfrac{\sqrt{(2\ m)^2+(4\ m)^2}}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}}, F 23 , x = − F 23 ⋅ r 13 r 23 = − 7.2 ⋅ 1 0 − 9 N ⋅ ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 ( 2 m ) 2 + ( 4 m ) 2 , F 23 , x = − 4.3 ⋅ 1 0 − 9 N . F_{23, x}=-4.3\cdot10^{-9}\ N. F 23 , x = − 4.3 ⋅ 1 0 − 9 N . F 23 , y = F 23 ⋅ r 12 r 13 = 7.2 ⋅ 1 0 − 9 N ⋅ 6 m ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 , F_{23,y}=F_{23}\cdot\dfrac{r_{12}}{r_{13}}=7.2\cdot10^{-9}\ N\cdot\dfrac{6\ m}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}}, F 23 , y = F 23 ⋅ r 13 r 12 = 7.2 ⋅ 1 0 − 9 N ⋅ ( 6 m ) 2 + ( ( 2 m ) 2 + ( 4 m ) 2 ) 2 6 m , F 23 , y = 5.77 ⋅ 1 0 − 9 N . F_{23,y}=5.77\cdot10^{-9}\ N. F 23 , y = 5.77 ⋅ 1 0 − 9 N . Then, we get:
F x = F 13 , x + F 23 , x = 7.2 ⋅ 1 0 − 10 N + ( − 4.3 ⋅ 1 0 − 9 N ) = − 3.58 ⋅ 1 0 − 9 N , F_x=F_{13,x}+F_{23,x}=7.2\cdot10^{-10}\ N+(-4.3\cdot10^{-9}\ N)=-3.58\cdot10^{-9}\ N, F x = F 13 , x + F 23 , x = 7.2 ⋅ 1 0 − 10 N + ( − 4.3 ⋅ 1 0 − 9 N ) = − 3.58 ⋅ 1 0 − 9 N , F y = F 13 , y + F 23 , y = ( − 9.66 ⋅ 1 0 − 10 N ) + 5.77 ⋅ 1 0 − 9 N = 4.8 ⋅ 1 0 − 9 N . F_y=F_{13,y}+F_{23,y}=(-9.66\cdot10^{-10}\ N)+5.77\cdot10^{-9}\ N=4.8\cdot10^{-9}\ N. F y = F 13 , y + F 23 , y = ( − 9.66 ⋅ 1 0 − 10 N ) + 5.77 ⋅ 1 0 − 9 N = 4.8 ⋅ 1 0 − 9 N . We can find the total electrostatic force on the charge q 3 q_3 q 3
F = F x 2 + F y 2 , F=\sqrt{F_x^2+F_y^2}, F = F x 2 + F y 2 , F = ( − 3.58 ⋅ 1 0 − 9 N ) 2 + ( 4.8 ⋅ 1 0 − 9 N ) 2 = 5.98 ⋅ 1 0 − 9 N . F=\sqrt{(-3.58\cdot10^{-9}\ N)^2+(4.8\cdot10^{-9}\ N)^2}=5.98\cdot10^{-9}\ N. F = ( − 3.58 ⋅ 1 0 − 9 N ) 2 + ( 4.8 ⋅ 1 0 − 9 N ) 2 = 5.98 ⋅ 1 0 − 9 N . We can find the direction of the total electrostatic force on the charge q 3 q_3 q 3
t a n θ = F y F x , tan\theta=\dfrac{F_y}{F_x}, t an θ = F x F y , θ = t a n − 1 ( F y F x ) , \theta=tan^{-1}(\dfrac{F_y}{F_x}), θ = t a n − 1 ( F x F y ) , θ = t a n − 1 ( 4.8 ⋅ 1 0 − 9 N − 3.58 ⋅ 1 0 − 9 N ) = 53. 3 ∘ . \theta=tan^{-1}(\dfrac{4.8\cdot10^{-9}\ N}{-3.58\cdot10^{-9}\ N})=53.3^{\circ}. θ = t a n − 1 ( − 3.58 ⋅ 1 0 − 9 N 4.8 ⋅ 1 0 − 9 N ) = 53. 3 ∘ . The total electrostatic force on the charge q 3 q_3 q 3 53. 3 ∘ 53.3^{\circ} 53. 3 ∘ x x x
#339914 on Dec 2023
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