A +4.0 uC charge is placed on the x axis at x= +3.0 m, and a -2.0 uC is located on the y-axis at y= -1.0 m. Point A is on the y axis at y= +4.0 m. Determine the electric potential at point A (relative to zero at the origin).

Answers

Answer 1

Answer:

The potential is  [tex]V_A = 9600 \ V[/tex]

Explanation:

From the question we are told that  

   The  magnitude of the charge is  [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]

   The position of the charge is  [tex]x = + 3.0 \ m[/tex]

   The magnitude of the second charge is  [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]

   The position is  [tex]y_1 = - 1.0 \ m[/tex]

     The position of point A is  [tex]y_2 = + 4.0 \ m[/tex]  

Generally the electric potential  at A due to the first charge is mathematically represented as

         [tex]V_a = \frac{k * q_1 }{r_1 }[/tex]

Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]

        [tex]r_1[/tex]  is the distance between first charge and a which is mathematically represented as

         [tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]

=>      [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]    

=>      [tex]r_1 = 5 \ m[/tex]    

So

           [tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]

      [tex]V_a = 7200 \ V[/tex]

Generally the electric potential  at A due to the second charge is mathematically represented as

         [tex]V_b = \frac{k * q_2 }{r_2 }[/tex]

Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]

        [tex]r_2[/tex]  is the distance between second charge and a which is mathematically represented as

         [tex]r_2 = y_2 - y[/tex]

=>      [tex]r _2 = 4.0 - (-1.0)[/tex]    

=>      [tex]r = 5 \ m[/tex]    

So

           [tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]

      [tex]V_a = -3600 \ V[/tex]

So the net potential difference at point A  due to the charges is mathematically represented as

       [tex]V_n = V_a + V_b[/tex]

=>     [tex]V_n = 7200 - 3600[/tex]

=>     [tex]V_n = 3600 V[/tex]

Generally the net potential difference at the origin due to both charges is mathematically represented as

     [tex]V_N = V_c + V_d[/tex]

Here  

      [tex]V_c = \frac{k * q_1 }{x}[/tex]

=>   [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]

=>   [tex]V_c = 12000 V[/tex]

and

              [tex]V_d= \frac{k * q_2 }{y}[/tex]

=>   [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]

=>   [tex]V_c =- 18000 V[/tex]

Generally the net potential difference at the origin is  

       [tex]V_N = 12000 - 18000[/tex]

=>     [tex]V_N = -6000[/tex]

Generally the potential difference at A relative to zero at the origin is mathematically evaluated as

         [tex]V_A = V_n - V_N[/tex]

=>      [tex]V_A = 3600 - (-6000)[/tex]

=>      [tex]V_A = 9600 \ V[/tex]


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Answers

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Explanation:

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Answers

Answer:

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Question 25
1 pts
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What is the correct free body diagram for the block of cheese? (help!!!)

Answers

Answer:

Explanation:

KHAN ACADEMY

Free body diagram of a block of cheese is sliding up an inclined surface that is not smooth is shown in the diagram.

What is free body diagram?

A graphic, dematerialized, symbolic representation of the body (a structure, element, or fragment of an element) in which all connecting "parts" have been eliminated is known as a free-body diagram.

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Then, free body diagram is according to the figure.

Learn more about free body diagram here:

https://brainly.com/question/24087893

#SPJ2

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answers

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  [tex]T_R = 11.8 \ days[/tex]  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  [tex]a_R = 45.0\ AU[/tex]

   The semi - major axis of  Planet D is  [tex]a_D = 60 \ AU[/tex]

    The orbital  period of planet D is  [tex]T_D = 18.164 \ days[/tex]

Generally from Kepler third law

          [tex]T \ \ \alpha \ \ a^{\frac{3}{2} }[/tex]

Here T is the  orbital period  while a is the semi major axis

So  

        [tex]\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}[/tex]

=>     [tex]T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }[/tex]  

=>     [tex]T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }[/tex]

=>      [tex]T_R = 11.8 \ days[/tex]  

   

The orbital period of the rocky debris in space is 12 days.

From Kepler's law, we know that the square of the period of a planet in years is equal to the cube of its mean distance from the sun in astronomical units (AU).

Mathematically;

[tex]T^2 = r^3[/tex]

Now, we have the mean distance as 45.0 AU, we need to obtain its period.

semi major axis of debris rR= 45.0 AU

semi major axis of planet D rD = 60 AU

Period of planet D  TD  = 18.164 days

Period of debris TR = ?

[tex]\frac{TR^2}{TD^2} = \frac{rR^3}{rD^3}[/tex]

[tex]\frac{rR^3}{rD^3} * TD^2[/tex]

[tex]TR^2 = 45^3/60^3 * 18.164^2[/tex]

TR = 12 days

Missing parts:

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

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