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
]
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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 = where k = coulomb's constant, q = charge and r = distance
Electric field intensity at point B,
= = 0.72 N/c now,
Electric field intensity at point A,
= = 0.36 N/c.
The angle between and is equal to 120°
Therefore, the resultant electric intensity at the corner C,
=
=
=
= 0.6235 N/C
Hence , The resultant electric intensity at the corner C is equal to 0.6235 N/C
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