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olchik [2.2K]
2 years ago
8

Three negative point charges q1 =-5 nC, q2 = -2 nC and q3 = -5 nC lie along a vertical line. The charge q2 lies exactly between

charge q1 and q3 which are 16 cm apart. Find the magnitude and direction of the electric field this combination of charges produces at a point P at a distance of 6 cm from the q2

Physics
1 answer:
Semmy [17]2 years ago
3 0

Answer:

Ep= 5450N/C in direction (-x)

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

1cm=  10⁻²m

Graphic attached

The attached graph shows the field due to the charges q₁, q₂ y q₃ in the  P  (x=6, y=0):

As the charge q₁ ,q₂ , and q₃ are negative the field enter the charges.

E₁: Electric Field due to charge q₁.  

E₂: Electric Field due to charge q₂.  

E₃: Electric Field due to charge q₃.

Field calculation due to q₁ and q₃

Because q₁ = q₃ and d₁ = d₃, then, the magnitude of E₁ is equal to the magnitude of E₃

d₁=d₃=d

d=\sqrt{6^{2}+8^{2}  } =10cm=0.1m

q₁=q₃= -5nC= 5*10⁻⁹C

E₁ = E₃= k*q/d² = 9*10⁹*5*10⁻⁹/0.1² = 4500 N/C

E₁x = E₃x= - 4500*(6/10)= -2700 N/C

E₁y =  -4500*(8/10)= -3600 N/C

E₃y=  +4500*(8/10)=+3600N/C

Field calculation due to q₂

E₂x= k*q₂/d₂²=  -9*10⁹*2*10⁻⁹/0.6²= -50 N/C

Magnitude and direction of the electric field in the point P (Ep)

Epx= E₁x+ E₂x+E₃x = -2700 N/C-50 N/C-2700 N/C= -5450N/C

Epy= E₁y+ E₃y= -3600 N/C+3600N  = 0

Ep= 5450N/C in direction (-x)

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1 year ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

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2 years ago
Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor
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156.8 Joules of energy is in the box's gravitational potential energy store

<u>Explanation</u>:

<em>Given:</em>

Mass of the box Dane is holding = 8 Kilograms

Height at which Dane is holding the box above the ground= 2 metres

<em>To Find:</em>

Gravitational potential energy in the box=?

<em>Solution:</em>

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PE_g=mgh

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h is the height

g is the gravitational force( 9.8 m/s^2)

Now substituting the given values,

PE_g=8\times 9.8\times 2

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A transition metal complex in solution has an absorption peak at 450 nm, in the blue region of the visible spectrum. What color
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Answer:

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Explanation:

The portion of UV-visible radiation that is absorbed implies that a portion of electromagnetic radiation is not absorbed by the sample and is therefore transmitted through it and can be captured by the human eye. That is, in the visible region of a complex, the visible color of a solution can be seen and that  corresponds to the wavelengths of light it transmits, not absorbs. The  absorbing color is complementary to the color it transmits.

So, in the attached image you can see the approximate wavelengths with the colors, where they locate the wavelength with the absorbed color, you will be able to observe the complementary color that is seen or reflected.

<u><em> In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).</em></u>

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