The magnitude of electric force on 2 micro coulomb charge placed at the centre o of two equilateral
Answers
Explanation:
First use Coulomb’s Law: F = kqQ / r^2 to calculate the forces between charges 1 and 3 and the force between charges 2 and 3.
Where the constant k = 8.99 x 10^9 Nm^2 / C^2
r is the distance between the charges
Calculate force charge 1 exerts on charge 3 making sure to convert the micro coulomb units to coulombs
F1,3 = (8.99 x 10^9) x (+6.3 x 10^-6) x (+6.3 x 10^-6) / (0.63)^2
F1,3 = 0.9 N
The positive sign means 1 is pushing 3 away from 1
Now calculate force charge 2 exerts on charge 3
F2,3 = (8.99 x 10^9) x (-6.3 x 10^-6) x (+6.3 x 10^-6) / (0.63)^2
F2,3 = -0.9 N
The negative sign means 2 is pulling 3 towards 2
Since 3 is pulling 2 at an angle to the X-axis and Y-axis, you will need to find the X and Y components of that force before combining it with the force 1 is exerting on 2.
F1,3x = 0.9 Cos 30⁰ = 0.779 N (right)
F1,3y = 0.9 Cos 60⁰ = -0.45 N (down)
F2,3x = -0.9 Cos 30⁰ = -0.779 N (left)
F2,3y = -0.9 Cos 60⁰ = -0.45 N (down)
Net Horizontal (X-axis) component = F1,3 x + F2,3x = 0.779 + (-0.779) = 0
Net Vertical (Y-axis) component = F1,3y + F2,3y = -0.45 + (-0.45) = -0.9 N
or 0.9 N down
Resultant force is calculated from combining these two force components
Resultant = SqRt(0 + 0.92) = SqRt(0 + 0.81) = SqRt (0.81) = 0.9 N down
SOLUTION⬆️
Refer to the attachment.
hope it helps ☺️
