Three point charges q_1, q_2, q_3 are placed | vedclass

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(B) For a system of charges,the net internal electrostatic force acting on the system is zero because the forces are action-reaction pairs.According t

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$(-3\hat{i} - 2\hat{j}) \, N$

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(B) For a system of charges,the net internal electrostatic force acting on the system is zero because the forces are action-reaction pairs.According to the principle of conservation of momentum or the property of internal forces in a closed system,the sum of all forces acting on the charges must be zero:$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$Given:$\vec{F}_1 = (2\hat{i} - \hat{j}) \, N$$\vec{F}_2 = (\hat{i} + 3\hat{j}) \, N$Substituting these values into the equation:$(2\hat{i} - \hat{j}) + (\hat{i} + 3\hat{j}) + \vec{F}_3 = \vec{0}$$(3\hat{i} + 2\hat{j}) + \vec{F}_3 = \vec{0}$Therefore,the force on $q_3$ is:$\vec{F}_3 = -(3\hat{i} + 2\hat{j}) \, N = (-3\hat{i} - 2\hat{j}) \, N$

Three point charges $q_1, q_2, q_3$ are placed at the vertices of a triangle. If the forces on $q_1$ and $q_2$ are $(2\hat{i} - \hat{j}) \, N$ and $(\hat{i} + 3\hat{j}) \, N$,respectively,then what will be the force on $q_3$?

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