Question

I need help with part A and part B. I tried those 3 answers and none were correct.

Answer 1

(A) It is given that (q ) = +2.6 uC
d = 31 cm = 0.31 m
Force on charge 2 due to charge 1
$F_{21} = \frac{kq_{2}q_{1}}{d^{2}}$
On putting the values we get
$F_{21} = \frac{9*10^{9}*2*2.6*10^{-6}*2.6*10^{-6}}{0.31^{2}}$
On solving we get
F₂₁ = 1.266 N
Now force on charge 2 due to charge 3
$F_{23} = \frac{kq_{2}q_{3}}{d^{2}}$
On putting the values we get
$F_{23} = \frac{9*10^{9}*2*2.6*10^{-6}*3*2.6*10^{-6}}{0.31^{2}}$
On solving , we get
F₂₃ = 3.798 N
Now force on charge 2 due to charge 4
$F_{24} = \frac{kq_{2}q_{4}}{(d\sqrt{2})^{2}}$
On putting the values we get
$F_{24} = \frac{9*10^{9}*2*2.6*10^{-6}*4*2.6*10^{-6}}{(0.31\sqrt{2})^{2}}$

On solving , we get
F₂₄ = 2.532 N
Now all forces are shown in the figure
In the x direction
F_X = -F₂₃ -F₂₄Cos45 = -3.798 - 2.532Cos45 = - 5.588 N
In the Y direction
F_Y = F₂₄Sin45 - F₂₁ = 2.532Sin45 - 1.266 = 0.524 N
Therefore the magnitude of the force would be
$F = \sqrt{F_{X}^{2} + F_{Y}^{2}}$
F = 5.613 N
Now the direction would be
$tan\theta = \frac{F_{Y}}{F_{X}}$
$tan\theta = \frac{0.524}{-5.588}$
$\theta = -5.357$ in clockwise direction from q₃ - q₂ direction
Now in counterclockwise direction it would be
$\phi$ = 180 + $\theta$
$\phi$= 174.64 ^o in counter clockwise direction from q₃ - q₂ direction

24 F, Sin45 24 Sin45 g-2q 43-3 q 45 23 F Cos45 21 45 44-4q