Question

ABC is an equilateral triangle of side 3cm charges of 5microcoulomb and -5microcoulomb are placed at A
and B find the electric field intensity at C​

Answer 1

☆Concept:

》Electric field lines emitting are always:

• comes towards the negative charge(-).
• moves away from the positive charge(+).

》At any point the net electric field(E) experienced will be the resultant of all field lines interacting at that point.

☆Formula:

$\boxed{\bf{ E = \dfrac{Kq}{r²}}}$

☆Solution:

see the solution with refer to above figure!

• at point C, we can see that the field lines of A and B are forming 60° with each other.

• let's solve it by vectors method;

• $\sf E_{net} = \sqrt{( \dfrac{Kq}{r²})² + ( \dfrac{Kq}{r²})²+ 2( \dfrac{Kq}{r²})( \dfrac{Kq}{r²})cos120°}$

• $\sf E_{net} = \sqrt{( \dfrac{Kq}{r²})² + ( \dfrac{Kq}{r²})²+ 2( \dfrac{Kq}{r²})( \dfrac{Kq}{r²})× \dfrac{-1}{2}}$

•$\sf E_{net} = \sqrt{( \dfrac{Kq}{r²})² + ( \dfrac{Kq}{r²})²– ( \dfrac{Kq}{r²})²}$

• $\sf E_{net} = \sqrt{( \dfrac{Kq}{r²})²}$

•$\sf E_{net} = \dfrac{Kq}{r²}$

(k=9×10⁹, $\sf q=5\mu C= 5×10^{-6}C$)

》$\sf E_{net} = \dfrac{9×10⁹(5×10^{-6})}{(3×10^{-2})²}$

》$\bf E_{net}= 5×10⁷N/C$

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