Physics: Three point charges lie in the xy-plane: with q1= 86 μc at origin (0,0), q2= 32μc at the point (2.5m, 0), and q3=−53μc at the point (1.5m, 2.2m). find the net force on q1?

Three point charges lie in the xy-plane: with

ayeofaygo5168

11.03.2022

$F_{net} = -0.7 \hat i + 4.77 \hat j$

Explanation:

Force due to q2 on q1 is along - X direction due to repulsion between them

so we have

$F_{12} = \frac{kq_1q_2}{r^2}$

$F_{12} = \frac{(9\times 10^9)(86 \mu C)(32 \mu C)}{2.5^2}$

$F_{12} = 3.96 N (-\hat i)$

Now force between q1 and q3 is given as

$F_{13} = \frac{kq_1q_3}{r^2} \frac{(1.5\hat i + 2.2 \hat j)}{\sqrt{1.5^2 + 2.2^2}}$

$F_{13} = \frac{(9\times 10^9)(86 \mu C)(53 \mu C)}{(1.5^2 + 2.2^2} \frac{(1.5\hat i + 2.2 \hat j)}{\sqrt{1.5^2 + 2.2^2}}$

$F_{13} = (5.78)\frac{(1.5\hat i + 2.2 \hat j)}{2.66}$

$F_{13} = (2.17)(1.5\hat i + 2.2 \hat j)$

Now net force on q1 is given as

$F_{net} = F_{12} + F_{13}$

$F_{net} = 3.96(-\hat i ) + (3.26 \hat i + 4.77 \hat j)$

$F_{net} = -0.7 \hat i + 4.77 \hat j$

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