Question

Two charges 15nC and 20nC placed at vertex of an equilateral triangle of side 5cm
What is the electric field at the third vertex?
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Answer 1

Explanation:

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Answer 2

Answer:

  E = 7(√3/10)K×10⁻⁹ v/m  

Explanation:

Given that ,

Two charges 15nC and 20nC are placed at vertex of an equilateral triangle of side 5cm.

To find the electric field at the third vertex of the equilateral triangle.

So, by using the formula of electric field (E)

         E = $\frac{kQ}{R^{2} }$

where Q is charge and R is distance between charge and point.

Electric field by charge 15 nC,

         E1 =  (k × 15 )/ (5)²

         E1 = 15K/25

         E1 = 3K/5

Electric field by charge 20 nC,

         E2 =  (k × 20 )/ (5)²

         E2 = 20K/25

         E2 = 4K/5

But the final electric field will be sum of E1cos30° and E2cos30°.

Because, the angle between E1 and E2 are 60° but final angle will be mid of both therefore it is 30°.

So,

     E = E1cos30° + E2cos30°

     E =  (3k/5)cos30° + (4k/5)cos30°

     E = kcos30° ( 3/5 + 4/5 )

     E = K (√3/2) (7/5)

     E = 7(√3/10)K×10⁻⁹ v/m       { 1nC = 10⁻⁹ }

THANKS.

#SPJ3.

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