Question

Chapter 21, Problem 062 In the figure what are the (a) magnitude and (b) direction (from x-axis in the counterclockwise direction) of the net electrostatic force on particle 4 due to the other three particles? All four particles are fixed in the xy plane, and q1 = 52.80 10-19 С, q2 . +52.80x 10-19 C, a:" + 105.60 x 10-19 С, a.-+ 52.80 x 10-19 C, 81-40°, d1-3.80 cm, and d2 - ds 4.30 cm. d. (a) Number Units (b) Number Units

Answer 1

Force between any two charges can be calculated using Coulomb's law I.e. $\fn_cm \large F = \frac{k|q1||q2|}{d^{2}}$ where d is the distance between two charges and k = 9 × 10⁹ Nm²C^-2. We will first calculate the forces F₁₄, F₂₄, F₃₄ their x and y components and add them respectively, we get F_x and F_y.

To calculate resultant force we will use the formula $\fn_cm \large F = \sqrt{F_{x}^{2}+F_{y}^{2}}$

To calculate direction we first calculate rhe angle $\fn_cm \large \Phi$ between F_x and F_y using formula $\fn_cm \large \tan \Phi = \frac{F_{y}}{F_{x}}$ then after calculating $\fn_cm \large \Phi$ we will add 180 degree to it to get the direction in counterclockwise from x axis

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