What is the net electrostatic force (magnitude and direction) on a particle with charge +5μc situated at the apex of an equili! ateral triangle if each of the other corners contain identical charges of-6 μc and the length of a side of the triangle is 0.10 m?

net electrostatic force = 46.76 N

and direction is vertical downward

Explanation:

given data

charge =  +5μC

corners identical charges = -6 μC

length of side = 0.10 m

to find out

What is the net electrostatic force

solution

here apex is the vertex where the two sides of equal length meet

so upper point A have charge +5μC and lower both point B and C have charge -6 μC

so

force between +5μC and -6 μC is express as

force  =       ..............1

put here electrostatic constant  k = 9 × Nm²/C² and q1 q2 is charge given and r is distance 0.10 m

so

force  =

force = 27 N

so net force is vector addition of both force

force =

here x is force 27 N

force =

net electrostatic force = 46.76 N

and direction is vertical downward

gravity

explanation:

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle