Three equal charges of +5.0 mu C each are held fixed at the three corners of an equilateral triangle of side 10cm . The coulomb force experienced by one of the charges due to the remaining two charges is
Answers
Given :
Three equal charges of +5.0 mu C each are held fixed at the three corners of an equilateral triangle of side 10cm
To Find :
Force on any one charge due other two charges .
Solution :
•Three charges are placed at corners of equilateral triangle of side 10 cm
=> Distance between charges (r) = 10 cm = 10^-¹ m
•Now
According to coulomb's law
F12 = Kq1q2/r² = KQ²/r² = F ___(say)
Where, F12 is Force on 1 due to charge 2
q1 & q2 are magnitude of charges
r is distance between charges
•Similarly ,
F13 = Kq1q3/r² = KQ²/r² = F
•Now , Angle between F12 & F13 is 60°
F²net = F12² + F13² +2F12.F13.cos60
F²net = F² + F² + 2F.F(1/2)
F²net = F² + F² + F²
F²net = 3F²
Fnet = √3F
Fnet = √3KQ²/r²
= √3 ×9×10^9(5×10^-6)²/(10^-¹)²
= √3 × 9×25 ×10^9×10^-¹²/10^-²
= √3 × 225 × 10¹¹×10-¹²
= √3×225/10
= 38.97 N
•Hence Force on any one charge due other two charges is 38.97 N