Question

1a0e Homework #1 problem #2 April 2 2020 Here are a few homework proplems. Find the net (magnitude & direction) force on 9,. Our first quiz will look Very similar, so this is good practice too! q=-600C +9= 500E Чm 4m 92 = - 2006 93 = -sone 19 = sono 4m Pq = -20cl Hints: * Opponit charges attrart #93 and 92 do push on each other, and you could calculate that, but we are focused on 9, only

Answer 1

Given :

q1 = 50 nC = 50 x 10-9 C

q2 = -20 nC = - 20 x 10-9 C

q3 = - 60 nC = - 60 x 10-9 C

Electrostatic force between two point charge = F = $\frac{kq_{1}q_{2}}{r^{2}}$ [where k = 9 x 109 Nm2C-2 , r = distance between the charges]

I will not use the sign of charge, as I have shown the direction of force.

a) see the diagram :

11 -9m v6 117

[force on q1 by q3 ]

k x 50 x 10-9 x 60 x 10-9 19193 =- i=- X OXIU A F

$\vec{F_{2}}=-\frac{kq_{1}q_{2}}{r^{2}}\hat{i} = -\frac{k\times 50\times 10^{-9}\times 20\times 10^{-9}}{4^{2}}\hat{j}$ [force on q1 by q2 ]

k x 50 x 10-9 x 20 x 10-9 k x 50 x 10-9 x 60 x 10-9 Fnet = F1+F2= -2 -(1687.5i + 562.5j) x 10-°N 42

Magnitude of net Force = $F_{net} = 1778.78 \times 10^{-9}N$

Direction of net Force : $\theta = tan^{-1}\frac{562.5}{1687.5} = tan^{-1}(\frac{1}{3}) = 18.43^{o}$ with -ve x-axis in 4th quadrant.

b) see the diagram :

Зm Y-um

r1 = 5 m

$\vec{F_{1}}=\frac{kq_{1}q_{3}}{r_{1}^{2}}(-cos37^{o}\hat{i}+sin37^{o}\hat{j}) = \frac{k\times 50\times 10^{-9}\times 60\times 10^{-9}}{5^{2}}(-0.8\hat{i}+0.6\hat{j})$    [force on q1 by q3 ]

$\vec{F_{2}}=-\frac{kq_{1}q_{2}}{r^{2}}\hat{i} = -\frac{k\times 50\times 10^{-9}\times 20\times 10^{-9}}{4^{2}}\hat{j}$   [force on q1 by q2 ]

$\vec{F_{net}} = \vec{F_{1}}+\vec{F_{2}}=\frac{k\times 50\times 10^{-9}\times 60\times 10^{-9}}{5^{2}}(-0.8\hat{i}+0.6\hat{j}) -\frac{k\times 50\times 10^{-9}\times 20\times 10^{-9}}{4^{2}}\hat{j} =(-864 \hat{i}+85.5\hat{j}) \times 10^{-9}N$

Magnitude of net Force = $F_{net} = 868.22 \times 10^{-9}N$

Direction of net Force : $\theta = tan^{-1}\frac{85.5}{864} = tan^{-1}(0.1) = 5.65^{o}$ with -ve x-axis in the 3rd quadrant.