Physics, asked by kaduatharva5, 9 months ago

A charge of √2C is placed at the top of your school building and another equal

charge at the top of your house. Take the separation between the two charges to be 2

km. How much force is exerted by the charges on each other?

Answers

Answered by abhi569
0

Explanation:

Using Coulumb's law.

     F = k q₁q₂/r²   k(for air)= 9*10⁹

F = k (√2 * √2)/(2 km)²

  = k * 2/(2000 m)²

  = 9 * 10⁹ * 2/4*10⁶  N

  = 9 * 10⁽⁹⁻⁶⁾ /2

  = 4.5 * 10³ N

Answered by VikalPandey
0

Answer:

4.5 kN

Explanation:

Both charges are equal

therefore, Q1 =Q2= √2

distance between the charges,

r = 2km

= 2000m { we can write it as 2x10^3 m }

k= 9x10^9 N

Now, by coulomb's law

F= k Q1 Q2/r^2

Put values

F= 9x10^9x√2x√2 / (2x10^3)^2

F= 9x10^9x2 / 4 x 10^6

F= 9x10^(9-6) / 2

F= 9x10^3 / 2

F= 4.5x10^3 N

Therefore the answer is 4.5x10^3 N or 4.5 kN

Thank you, Hope you understood this

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