Three point charges are positioned as follows: q₁ is at (0.00 m,0.00 m),q₂ is at (1.20 m,0.00 m),and q₃ is at (1.20 m,1.60 m).If q₁ = 1.00 $ \mu$C,q₂ = 2.00 $ \mu$C,and q₃ = 3.00 $ \mu$C,what is the magnitude of the force on q₂? A) 2.45 $\times$ 10^-2 N B) 1.25 $\times$ 10^-2 N C) 2.11 $\times$ 10^-2 N D) 3.36 $\times$ 10^-2 N E) 5.08 $\times$ 10^-2 N

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Quizplus · Accepted Answer

The force on q2 is due to the electric fields of q1 and q3. The electric field of q1 is in the -x direction and has a magnitude of:$$E_1 = k\frac{q_1}{r_1^2} = (9	imes10^9 \frac{Nm^2}{C^2})\frac{1	imes10^{-6} C}{(1.2 m)^2} = 6.25	imes10^3 N/C$$The electric field of q3 is in the -y direction and has a magnitude of:$$E_3 = k\frac{q_3}{r_3^2} = (9	imes10^9 \frac{Nm^2}{C^2})\frac{3	imes10^{-6} C}{(1.6 m)^2} = 3.52	imes10^3 N/C$$The net electric field on q2 is:$$E = E_1 + E_3 = (6.25	imes10^3 N/C)\hat{i} + (3.52	imes10^3 N/C)\hat{j}$$The force on q2 is:$$F = qE = (2	imes10^{-6} C)(6.25	imes10^3 N/C)\hat{i} + (2	imes10^{-6} C)(3.52	imes10^3 N/C)\hat{j}$$$$F = (1.25	imes10^{-2} N)\hat{i} + (7.04	imes10^{-3} N)\hat{j}$$The magnitude of the force is:$$|F| = \sqrt{(1.25	imes10^{-2} N)^2 + (7.04	imes10^{-3} N)^2} = 1.45	imes10^{-2} N$$

Three Point Charges Are Positioned as Follows: Q1 Is at (0.00

Three Point Charges Are Positioned as Follows: Q₁ Is at (0.00