charges of +5µc+10 μc and -10 µc are placed in air at the corners ab and c of an equilateral triangle abc having each side equal to 5 cm. determine the resultant force on the charge at a.

Asked by Sivanigopesh | 26th Apr, 2021, 03:24: PM

Expert Answer:

Figure shows Forces acting on +5μC charge placed at vertex A.
Force F1 is repulsive force between +5μC charge and +10μC charge.
Force F2 is attractive force between +5μC charge and -10μC charge .
 
Force between two charges q1 and q2 that are separated by a distance d is given as
 
begin mathsize 14px style F space equals space K space cross times fraction numerator q subscript 1 q subscript 2 over denominator d squared end fraction end style
Where K = 1/(4πε0 ) = 9 × 109 N m2 C-2  , is Coulomb's constant
 
Magnniude of each force is same because product of charges is same and distance between charges also same.
 
Hence Resultant force R = 2 |F| cos60 =  ( 2 × 9 × 109 × 50 × 10-12 × cos60) /  ( 25 × 10-4 ) = 180 N

Answered by Thiyagarajan K | 26th Apr, 2021, 05:08: PM