Question

Particle A of charge 3.30 10-4 C is at the origin, particle B of charge -6.24 10-4 C is at (4.00 m, 0), and particle C of charge 1.06 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C?

Answer 1

Answer:

a) $E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) zero

Explanation:

a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:

$E_T=E_1+E_2$

E1: electric field of charge 1

E2: electric field of charge 2

It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:

$E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]$

r13: distance between charges 1 and 3

r12: charge between charges 2 and 3

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

Thus, you first calculate the distance r13 and r23, and also the angles:

$r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°$

Next, you replace the values of all parameters in order to calculate E1 and E2:

$E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}$

finally, you obtain for ET:

$E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.