## ASAP please! Three positive charges are arranged as shown in the diagram. Three charges at the corners of an isosceles triangle. The two cha

Question

ASAP please! Three positive charges are arranged as shown in the diagram. Three charges at the corners of an isosceles triangle. The two charges at the base are +5 microcoulombs and +9 microcoulombs and are separated by .32 m. The triangle is .42 m tall up to a charge +3 microcoulombs. The other two sides of the triagle are both .45 m. What is the magnitude and direction of the net electrical force acting on the +3 µC charge? magnitude: 1.75 N direction: 6.2° north of east magnitude: 1.75 N direction: 6.2° south of east magnitude: 1.86 N direction: 21° north of east magnitude: 1.86 N direction: 21° south of east

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5 months 2021-08-24T03:56:04+00:00 2 Answers 6 views 0

## Answers ( )

1. Answer: magnitude: 1.75 N
; direction: 6.2° north of east

Explanation:

It’s right

Fn = 1,78 [N]

β = 6,4⁰   north-east direction

Explanation:

As all charges are positive, forces between electric charges located in the cornes of the base of the triangle and the charge located in the top of the isoceles triangle, are repulsion forces. The east-west direction will depend on where the biggest charge is located, if Q₁ = 9 microC, is in the left end of the triangle base then force on charge Q₃ = 3 microC will have direction north-east, if the above mention configuration change (that means Q₁ is in the right end of the triangle base) then the force will have north-west direction. As in the answer solution, 1,75 [N] direction north-east is offer, we will choose the first configuration. According to that:

K = 1/4*π*ε₀       K  = 8,98*10⁹ [N*m²/C²]

The angles in the base of the triangle are equals (we call it α )

sinα = 0,42/045 = 0,93

α = 69⁰

cosα = 0,356

And  Q₃ = 3 mC     = 3*10⁻⁶ C

Q₁  = 9  mC  =  9*10⁻⁶ C

Q₂  = 5  mC  =  5*10⁻⁶ C

F₁ [Q₁,Q₃]  =  K*Q₁*Q₃/d² ⇒F [Q₁,Q₃] = 8,98*10⁻⁹ *  9*10⁻⁶ * 3*10⁻⁶/(0,45)²[N]

F₁ [Q₁,Q₃]  = 1,2 [N]  direction along the side of the triangle therefore

its components are:

F₁ₓ = F₁* cosα   ⇒ F₁ₓ = 1,2 * 0,356

F₁ₓ = 0,43 [N]

F₁y = F₁ *sinα   ⇒ F₁y = 1,2 * 0,93

F₁y = 1,16  [N]

Now

F₂ [Q₂,Q₃] = K*Q₂*Q₃/d² ⇒F [Q₂,Q₃] = 8,98*10⁻⁹ *  5*10⁻⁶ * 3*10⁻⁶/(0,45)²[N]

F₂ = 0,67 [N]

F₂ₓ  =  F₂* cosα   ⇒ F₂ₓ  = 0,67*0,356

F₂ₓ  = 0,24 [N]

F₂y  = F₂*sinα  ⇒ F₂y  = 0,67*0,93

F₂y  = 0,62 [N]

We must remember F₁ and F₂ are repulsion forces, then components over y-axis have the same sense (we must add then) and components over  x-axis  have to be subtracted,  in this case, the sense is determine by major force component (to the right) according to this:

Fy = F₁y * F₂y = 1,16 + 0,62

Fy = 1,78 [N]

And

Fₓ = F₁ₓ – F₂ₓ  =  0,43 – 0,24

Fₓ = 0,19

Net force module is:

Fn = √ (1,78)² + (0,19)²

Fn = 1,79  [N]

Having an angle respect the y-axis:

tan β = 0,19/1,78  ⇒  tan β = 0,11 and from tan  tables

β = 6,4⁰  north-west direction