Question

Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x=0.00 m,y=0.600 m) , and the second charge has magnitude of q 2 = 3.40 nC q2=3.40 nC and is located at ( x = 1.30 m , y = 0.400 m ) (x=1.30 m,y=0.400 m) , calculate the x x and y y components, E x Ex and E y Ey , of the electric field, → E E→ , in component form at the origin, ( 0 , 0 ) (0,0) . The Coulomb force constant is 1 / ( 4 π ϵ 0 ) = 8.99 × 10 9 N ⋅ m 2 /C 2 1/(4πϵ0)=8.99×109 N⋅m2/C2 .

Answer 1

Answer:

$\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

Explanation:

In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

$\theta_1=tan^{-1}(\frac{0.6m}{0m})=90\° \\\\\theta_2=tan^{-1}(\frac{0.4m}{1.3m})=17.10\°$

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

$\vec{E} = E_x \hat{i}+E_y\hat{j}\\\\\vec{E}=(E_1cos\theta_1-E_2cos\theta_2)\hat{i}+(E_1sin\theta_1-E_2sin\theta_2)\hat{j}\\\\E_1=k\frac{q}{r_1^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{2.75*10^{-9}C}{(0.6m)^2}=68.67\frac{N}{C}\\\\E_2=k\frac{q}{r_2^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{3.40*10^{-9}C}{((1.3m)^2+(0.4m)^2)}=16.52\frac{N}{C}$

Hence, by replacing E1 and E2 we obtain:

$\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

hope this helps!!