A4.89 μc test charge is placed 4.10 cm away from a large, flat, uniformly charged nonconducting surface. the force on the charge is 321 n. the charge is then moved 2.00 cm farther away from the surface. what is the force on the test charge now?

The new force on the test charge is 145.02N

Explanation:

Force on a unit positive charge can be calculated using coulomb's law.

F =Kq²/r²

Where

F is the force on the charge = 321 N

K is a constant = 8.99 X10⁹ Nm²/C²

q is a test charge = 4.89 μC  = 4.89 X10⁻⁶ C

r is the distance between the charge and the surface = 4.1cm

F ∝1/r²

Force on the charge is inversely proportional the square of distance between the charge and the surface.

Fr² = constant

F₁r₁² = F₂r₂²

F₂ = F₁r₁²/r₂²

If the charge is then moved 2.00 cm farther away from the surface;

The new distance becomes 4.10 cm + 2.00 cm = 6.1 cm

Therefore, r₂ = 6.1 cm, r₁ = 4.1 cm, F₁ =321 N

F₂ = (321 X4.1²)/(6.1²)

F₂ = 145.02N

Therefore, The new force on the test charge is 145.02N

the pull of gravity on the rocket decreases

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this