Question

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?

Answer 1

net electrostatic force = 46.76 N

and direction is vertical downward

Step-by-step 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 =
`k (q1q2)/(r^2)` ..............1

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

so

force =
`9*10^(9) (5*6*10^(-12))/(0.10^2)`

force = 27 N

so net force is vector addition of both force

force =
`\sqrt{x^(2)+x^(2)+2x^(2)cos60}`

here x is force 27 N

force =
`\sqrt{27^(2)+27^(2)+2(27)^(2)cos60}`

net electrostatic force = 46.76 N

and direction is vertical downward