Question

Please help ASAP.

Question 2 In the figure above, charge A is-1.65 nC, charge B is 3.30 nC, and charge C is 1.65 nC. If x-1.55 cm and y= 3.10 cm, what is the electric field at the dot? 7.93x104N/C, 38.9° clockwise from the positive xaxis 5.96x104 N/C, 38.9° clockwise from the positive xaxis 7.93x104 N/C, 19.4° counter-clockwise from the positive xaxis O 5.96×104 N/C, 19.4° clockwise from the positive x-axis

Answer 1

ANswer is D

Angle made by charge A

$\theta =tan^{-1}\left ( \frac{3.1}{1.55} \right )=63.435^{o}$

X-Component of Net electric field

$E_{x}=\frac{K|Q_{C}|}{x^{2}}-\frac{K|Q_{A}|}{\left ( \sqrt{x^{2}+y^{2}} \right )^{2}}Cos63.435$

E_x=(9*10⁹)[(1.65/0.0155²)-(1.65Cos63.435/(0.0155²+0.031²)]*10^-9

E_x=56282.11 N/C

Y-Component of net electric field

$E_{y}=-\frac{K|Q_{B}|}{y^{2}}+\frac{K|Q_{A}|}{\left ( \sqrt{x^{2}+y^{2}} \right )^{2}}Sin63.435$

E_y=(9*10⁹)[(-3.3/0.031²) +(1.65Sin63.435/(0.031²+0.0155²)]*10^-9

E_y=-19848.28 N/C

Magnitude

|E|=sqrt[(56282.11)²+(-19848.28)²] =5.96*10⁴ N/C

Direction

o=tan^-1(-19848.28/56282.11) =-19.4^o =19.4^o Clockwise from positive x-axis