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mamaluj [8]
3 years ago
9

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nitude and direction of the resultant electric force acting on a charge of 3.0 × 10−9 C located at x = 0.70m.
Physics
1 answer:
nadya68 [22]3 years ago
8 0

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β  =  0,5/0,7

tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

cos  β = 0,81

d = √ (0,5)²  +  (0,7)²       d1stance between charges

d = √0,25 + 0,49

d =  √0,74  m

d = 0,86  m

Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

q₁  =  2,5* 10⁻⁹C

q₂  = 3,0*10⁻⁹C

d²  = 0,74 m²

Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

And  Fx  =  F*cos  β

Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

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<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

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T=?

we have the formulas,

I = (1/6)Md2

F = ma

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k = mω2 τ = -d(FgsinΘ)

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Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

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<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the  torque value </u>

τ = -dmgsinΘ    where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

<u>Rotational to linear form</u>

a = αr  

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

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