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miv72 [106K]
3 years ago
11

A 7.0-μC point charge and a point charge are initially extremely far apart. How much work does it take to bring the point charge

to the point , and the point charge to the point ( k = 1/4 πε 0 = 9.0 × 10 9 N ∙ m 2/C 2)?
Physics
2 answers:
vazorg [7]3 years ago
4 0

Answer:1.008 ×10^-14/rJ

Where r is the distance from.which the charge was moved through.

Explanation:

From coloumbs law

Work done =KQq/r

Where K=9×10^9

Q=7×10^-6C

q=e=1.6×10^-19C

Micro is 10^-6

W=9×10^9×7×10^-6×1.6×10^-19/r=100.8×10^-16/r=1.008×10^-14/rJ

r represent the distance through which the force was used to moved the charge through.

PtichkaEL [24]3 years ago
4 0

Complete question:

A 7.0 -μC point charge and a 9.0 -μC point charge are initially infinitely far apart. How much work does it take to bring the 7.0-μC point charge to x=3.0 mm, y= 0.0mm and the 9.0-μC point charge to x=-3.0 mm, y=0.0 mm? (the value of k is 9.0*10^9 N*m^2/C^2)

Answer:

  • Work done in bringing 7.0-μC point charge from infinity to x=3.0 mm, y= 0.0mm is 21 x 10⁶ Volts
  • work done in bringing 9.0-μC point charge from infinity to x=-3.0 mm, y=0.0 mm is -27 x 10⁶ Volts.

Explanation:

Work done in bringing a unit positive charge from infinity to that point in electric filed (V) = Electric field strength X distance

V = EXd = \frac{kq}{d^2}Xd=\frac{kq}{d}

where;

K is a constant = 9X10⁹ N.m²/C²

q is point charge in C

d is the distance in m

--------------------------------------------------------------------------------------------

Work done in bringing 7.0-μC point charge from infinity to x=3.0 mm, y= 0.0mm

V = \frac{(9X10^9) X(7X10^{-6})}{3x10^{-3}}

V = 21 x 10⁶ Volts

Work done in bringing 9.0-μC point charge from infinity to x=-3.0 mm, y=0.0 mm

V = \frac{(9X10^9) X(9X10^{-6})}{-3x10^{-3}}

V = -27 x 10⁶ Volts

--------------------------------------------------------------------------------------------

Work done in bringing 7.0-μC point charge from infinity to x=3.0 mm, y= 0.0mm is 21 x 10⁶ Volts While work done in bringing 9.0-μC point charge from infinity to x=-3.0 mm, y=0.0 mm is -27 x 10⁶ Volts.

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(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
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a) t = -\frac{ln(2)}{k}

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A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

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Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

\frac{1}{2} = e^{kt}

Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

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And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

For this case we want to find the value of t when we have remaining \frac{A_o}{8}

So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

\frac{1}{8} = 2^{-\frac{t}{T}}

We can apply natural log on both sides and we got:

ln(\frac{1}{8}) = -\frac{t}{T} ln(2)

And if we solve for t we got:

t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

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Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

8 0
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The new period is D) √2 T

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

\texttt{ }

\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}

T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}

T : T_2 = \sqrt{L} : \sqrt{2L}

T : T_2 = 1 : \sqrt{2}

\boxed {T_2 = \sqrt{2}\ T}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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