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Softa [21]
3 years ago
14

What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Physics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0

Complete Question

The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Answer:

The  value  is  P = 3.270960 *10^{6} \  W

Explanation:

From the question we are told that

   The  electric field strength is  E  = 4.0 \  M \  V/m  =  4.0 *10^6 \  V/m

    The  diameter is  d =  1.4 \  cm  = 0.014 \ m

Generally the radius is mathematically represented as

        r = \frac{d}{2}

=>     r = \frac{0.014}{2}

=>    r   = 0.007 \  m

Generally the cross-sectional area is mathematically represented as

        A =  \pi r^2

        A =  3.142 *  (0.007)^2

       A = 0.000154 \ m^2

Generally the maximum power that can be delivered is mathematically represented as

        P = \frac{c *  \epsilon_o  *  E^2 *  A }{2}

Here c is the speed of light with value  c =  3.0*10^{8} \  m/s

        \epsilon_o is the permittivity of free space with value  \epsilon_o  =  8.85 *10^{-12}  \ m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

     P =  \frac{3,0*10^8 *  8.85*10^{-12} *  (4 *10^6)^2 * 0.00154}{ 2}

       P = 3.270960 *10^{6} \  W

         

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The force between a pair of .005 charges is 750 N. What is the distance between them?
Ganezh [65]

Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

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Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

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