Physics: Charge q1 = 6.0 nc is at (0.30 m, 0), charge q2 = -1.0 nc is at (0, 0.10 m), and charge . what are the magnitude and direction of the net electrostatic force on the 5.0-nc charge due to the other charges?

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Charge q1 = 6.0 nc is at (0.30 m, 0), charge q2 = - id160030021718109, lesliealvarado1022

benbeltran9030

16.03.2022

$|F_3| = 5.4*10^{-6}N$

$Direction 56.31^{0}$ W from -ve x-axis

Explanation:

electrostatic force is given by

$F =\frac{k*Q1*Q2}{R^2}$

if both charge have different signs, Force will be attractive, and if both charge have same signs, force will be repulsive .

Now Net force will be

force between q1 and q3 is repulsive and towards -ve x-axis, which is

where q3 = 5.0nC {GIVEN}

$F_13 =\frac{k*q1*q3}{r_{13}^2} (-i)$

force between q2 and q3 is attractive and towards +ve y-axis, which is

$F_23 = \frac{k*q2*q3}{r_{23}^2} j$

Now given that

q_1 = 6 nC, q_2 = -1 nC, q_3 = 5.0 nC

r_13 = 0.30 m

r_23 = 0.10 m

total force will be

$F_3 = F_{13} (-i) + F_{23} (j)$

F_3 = $\frac{9*10^9*6*10^{-9}*5*10^{-9}}{0.30^2} (-i) +\frac{9*10^9*1*10^{-9}*5*10^-{9}}{0.10^2} j$

$F3 = -3*10^{-6}i + 4.5*10^{-6}j N$

magnitude of force can be calculated as

$|F3| = \sqrt{(- 3*10^{-6})^2 + (4.5*10^-6)^2}$

$|F_3| = 5.4*10^{-6}N$

Direction $= tan^{-1}\frac{Fy}{Fx} = tan^{-1}\frac{4.5}{3} = 56.31^{0}$ W from -ve x-axis

Direction = 180 - 56.31 = 123.69 anti- Clockwise from +ve x-axis

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