How Do You Calculate the Net Electric Force on a Particle?

[jelliDollFace]

Homework Statement

Particle B, with positive charge q_a, fixed in the yz-plane at (0,d_2,d_2). What is the net force F on particle A (positive) due to this charge?
Express your answer (a vector) using k, q_0, q_3, d_2, x hat, y hat, and z hat. Include only the force caused by particle B.

Homework Equations

electric force F = k(q_a)(q_b)/(d_2)^2
k = 9*10^9
q_a,q_b = point charges
d_2 = distance between charges

The Attempt at a Solution

i really not sure how to go about determining the components of the force due to particle B
could someone help me with some guidelines on how to do so pls or a link to somewhere, couldn't find too much when i searched

do i use the z and x distances to form a triangle and then use the side lengths to determine the angle theta?

thanks so much

 

[LowlyPion]

Homework Statement

Particle B, with positive charge q_a, fixed in the yz-plane at (0,d_2,d_2). What is the net force F on particle A (positive) due to this charge?
Express your answer (a vector) using k, q_0, q_3, d_2, x hat, y hat, and z hat. Include only the force caused by particle B.

Homework Equations

electric force F = k(q_a)(q_b)/(d_2)^2
k = 9*10^9
q_a,q_b = point charges
d_2 = distance between charges

The Attempt at a Solution

i really not sure how to go about determining the components of the force due to particle B
could someone help me with some guidelines on how to do so pls or a link to somewhere, couldn't find too much when i searched

do i use the z and x distances to form a triangle and then use the side lengths to determine the angle theta?

thanks so much

Coulomb's law is based on the distance between the charges.

What is the distance if the charge is located d₂ y and d₂ z away?

 

[jelliDollFace]

it would just be the hypotenuse using d2_y and d2_z as the sides, so distance = sqrt(d2_y^2 + d2_z^2), so is the triangle formed a 45-45-90?

 

[LowlyPion]

it would just be the hypotenuse using d2_y and d2_z as the sides, so distance = sqrt(d2_y^2 + d2_z^2), so is the triangle formed a 45-45-90?

Yes, making the distance effectively then (2)^1/2 * d₂

 

[jelliDollFace]

so using the constants stated in problem the force would be:

F = (k*q_0*q_3)/ (d_2(2^1/2))^2, how would i state the unit vectors the problem involves components, i know it involves x hat and z hat?

 

[LowlyPion]

so using the constants stated in problem the force would be:

F = (k*q_0*q_3)/ (d_2(2^1/2))^2, how would state the unit vectors the problem involves components, i know it involves x hat and z hat?

$$\vec{F} = 0 \hat{x} + \frac{F}{\sqrt{2}}\hat{y} + \frac{F}{\sqrt{2}}\hat{z}$$

Where

$$F = \frac{k*q_0*q_3}{(d_2* \sqrt{2})^2}$$

 

[jelliDollFace]

ohhh, but why is F divided by 2? i must've missed something

 

[LowlyPion]

ohhh, but why is F divided by 2? i must've missed something

Because those are the components of the force in each direction.

But oops that should be over the sqrt of 2. Sorry.

The Root Sum of the Squares = the F

Edit: I just changed it.

 

[LowlyPion]

In actual practice that would be the Cos or Sin of the angle contribution of the vector components.

It so happens in this case that the angles are Sin45 and Cos45 and = 1/2*sqrt(2) or 1/sqrt(2)

 

[jelliDollFace]

thanks so much, actually the force components were supposed to be negative since the two point charges were of the same charge and thus repelled each other, so in other words replace the + with -

 

[LowlyPion]

thanks so much, actually the force components were supposed to be negative since the two point charges were of the same charge and thus repelled each other, so in other words replace the + with -

Yes, that's correct. The question was with respect to forces at A.