Question

Two point charges of 2nC and -2nC are placed at 2 corners of an equilateral triangle of side 5 cm. Find the magnitude of the resultant electric field at the third corner

Answer 1

Explanation:

the magnitude of resultant electric field at third corner is 7200 nanocoulomb

Answer 2

Answer:

The resultant electric field at the third corner is 7200N/C.

Explanation:

Let the two point charges on each corner of the equilateral triangle be $q_{1}$ and $q_{2}$ respectively.

Let the 3 corners of the triangle be A, B and C with the side of each 5cm. It should also be known that in an equilateral triangle, each interior angle of a triangle is 60°.

The electric field at C is due to the point charge 2nC at A

Therefore, the resultant electric field at the third corner C is given by,

E = $\frac{2kq}{r^{2} }$cos60°  , where k = 1/4πε , $q_{1}$ = 2 x $10^{-9}$ C and r = 0.05m

Therefore, on substituting the values of k, $q_{1}$ and r we get,

E = $\frac{2*2*10^{-9} *1}{0.005^{2}*4*3.14*2*8.854*10^{-12} }$

  =  $\frac{2*10^{3} }{0.0025*4*3.14*8.854}$

  =  $\frac{2*10^{3} }{12.56*0.0025*8.854}$

  =  $\frac{2*10^{3} }{0.0314*8.854}$

  =   $\frac{2*10^{3} }{0.2780}$

  =   7.1938 x 10³ N/C

  =  7193.8 N/C

which can be approximated as 7200N/C.

Therefore, the resultant electric field at the the third corner is 7200 N/C.

The pictorial representation of the forces of the charges acting upon each other is attached in the image.

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