What must the charge (sign and magnitude) of a particle of mass 1.47 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 680 n/c ? use 9.81 m/ s 2 for the magnitude of the acceleration due to gravity. view available hint(s) nothing c c submit part b what is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? use 1.67×10−27 kg for the mass of a proton, 1.60×10−19 c for the magnitude of the charge on an electron, and 9.81 m/ s 2 for the magnitude of the acceleration due to gravity. view available hint(s) nothing n/c n / c submit provide feedback next

Part a)

Part b)

Explanation:

Part a)

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force

so here we have

and its sign must be negative so that it will have upward electric force

so it is

Part b)

as we know that weight of proton is balanced by electric force

so we will have

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)