Question

ABC is an equilateral triangle of side 0.1 m. Charges of 4 nC and -8 nC

are placed at the corners A and B respectively. Calculate the resultant electric intensity at the corner C.

[E = 6235.38NC-1

]​

Answer 1

The resultant electric intensity at the corner C is equal to 0.6235 N/C

Given that;

ABC is an equilateral triangle of side 0.1 m. Charges of 4 NC and -8 NC are placed at corners A and B respectively

To find;

The resultant electric intensity at the corner C.

Solution;

We know that the electric field intensity is given by the formula,

E = $\frac{Kq}{r^{2} }$  where k = coulomb's constant, q = charge and r =  distance

Electric field intensity at point B,

$E_{B}$ = $\frac{9 X 10^{9} X 8 X 10^{-9} }{10^{-2} }$ = 0.72 N/c now,

Electric field intensity at point A,

$E_{A}$ =  $\frac{9 X 10^{9} X 4 X 10^{-9} }{10^{-2} }$ = 0.36 N/c.

The angle between $E_{A}$ and $E_{B}$ is equal to 120°

Therefore,  the resultant electric intensity at the corner C,

=  $\sqrt{E_{A} ^{2} + E_{B} ^{2} + 2E_{A}E_{B}Cos120 }$

= $\sqrt{(0.72)^{2} + (0.36)^{2} + 2 X 0.72 X 0.36 X -1/2}$

= $\sqrt{0.3888}$

= 0.6235 N/C

Hence , The resultant electric intensity at the corner C is equal to 0.6235 N/C

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