1
$\begingroup$

Four charged particles are arranged in a square as shown below, with A=3, B=4, and C=5

enter image description here

(a) Determine the electric field at the location of charge q. (Use the following as necessary: $q$, $a$, and $k_e$.) Both the magnitude and the direction, with direction given in degrees counterclockwise from the x-axis.

(b) Determine the total electric force exerted on q. (Use the following as necessary: $q$, $a$, and $k_e$.) Again, direction and magnitude.

The work I have so far:

I know that $E=\frac{k_eq}{r^2}$, so I figured that $E_A=\frac{3qk_e}{a^2}$ and $E_C=\frac{5qk_e}{a^2}$ and $E_B=\frac{4qk_e}{2a^2}$ since the distance between B and the upper right corner is $\sqrt{2a^2}$, and I think I have to split $E_B$ into the x and y components to add them, but how does that give me a single magnitude and not a combination of x and y parts? I honestly have no idea how to find the direction of the field.

part b should be similar, except with $F_e=\frac{k_eq_1a_2}{r^2}$ for each one, which I have worked out, and then added similarly with B split into x and y components, right? Again, not sure the formula or trick to find the direction of the force.

$\endgroup$
3
  • $\begingroup$ This might be the confusing part: how do you get the x and y components of $E_B$? $\endgroup$ Commented Jun 8, 2016 at 22:07
  • $\begingroup$ @philip_0008, wouldn't the x be $E_Bcos(45)$ and y be $E_Bsin(45)$? $\endgroup$
    – Amanda
    Commented Jun 8, 2016 at 22:10
  • $\begingroup$ Ah you're right. Even that confuses me. I've just deleted an answer saying use $a$ for the distance of x-component:) $\endgroup$ Commented Jun 8, 2016 at 22:12

2 Answers 2

1
$\begingroup$

If you found the total x and y component of the total field just use:

Magnitude E_{tot}:

$$E_{tot}=sqrt(E^2_{tot-x}+E^2_{tot-y})$$

The direction in angle:

$$\theta = arctan(\frac{E_{tot-y}}{E_{tot-x}})$$

The magnitude of force would just be $F=qE$

Note: you should be careful in angles as it may sometimes not the angle measured from +x as the norm, for you to detmine it, check at which quadrant your resultant vector lie, Its for you to figure out (Trigo basics).

$\endgroup$
1
$\begingroup$

a) Add the x and y components separately, then calculate the magnitude E (the resultant field) and the angle it makes with the x axis.

b) The force is qE, in the same direction as E.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.