Question

Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at (LL) Q3 is positive and at (4,0) Q4 is negative and at (0,0) A) Draw and label a diagram of the described arrangement described above (include a coordinate system). B) Determine the force that charge Q1 exerts on charge Qz. C) Determine the force that charge Q3 exerts on charge Q2. D) Determine the force that charge Q4 exerts on charge Q2. E) Now assume that all the charges have the same magnitude (Q) and determine the net force on charge Q2 due to the other three charges. Reduce this to the simplest form (but don't put in the numerical value for the force constant).

Answer 1

Answer:

A) See Annex

B) Fq₁₂ = K *  Q₁*Q₂ /16 [N] (repulsion force)

C)  Fq₃₂  = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q²  [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L²    ⇒  d² = 2*L²     ⇒ d = √2*L²   ⇒ d= (√2 )*L

d = 4√2 units of length   (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂  = K * Q₁*Q₂ /(L)²     Fq₁₂ = K *  Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂  = K * Q₃*Q₂ /(L)²     Fq₃₂  = K * Q₃*Q₂ /16  (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)²      Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that  Q₄ exerts on Q₂  and Q₂ = Q  ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32       F₄₂  = K* Q²/32  [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2   and so is F₄₂(y)  =  [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces  Fq₁₂  and

Fq₃₂  respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

In new conditions

Fq₁₂ = K *  Q₁*Q₂ /16    becomes  Fq₁₂ = K * Q²/ 16 [N]   and

Fq₃₂ = K* Q₃*Q₂ /16      becomes   Fq₃₂ = K* Q² /16  [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x)  = Fnx

and over y-axis, we subtract   Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2  ⇒  Fnx =  K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude  then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces