Explanation:
use equation power=watts/time
to find time rearrange to make time = power/wats
so you have your equation substitute the numbers
so 9560J/860W is 11 minutes
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer: The initial force is reduced a factor 1/4 when the separation between charge is doubled
Explanation: As it well known the electric force between two charges is given by:
Finitial=k*q1*q2/d^2 where d is the distance between charges and k is a constant
if the distance is doubled this means 2*dinitial thus the new force is equal to F initial* 1/4
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J