Acapacitor with plates separated by distance do is charged to a voltage difference δvc. all wires and batteries are disconnected, then the two plates are pushed together (with insulated handles so charges cannot flow) to a new separation distance that was half the original distance ( ). which of the following quantities will be doubled as a result of this new separation distance?
a. the capacitance of the capacitor, c.
b. the energy stored in the capacitor, u.
c. the magnitude of the electric field between the plates, ∣ e -> ∣.
d. the voltage drop across the capacitor, δvc
e. all of the above choices are correct.

a. The capacitance of the capacitor, C.

Explanation:  

Applying Gauss´Law to a closed surface enclosing one of the capacitor plates, it can be showed that the capacitance of a parallel plates capacitor, depends only on his geometry and the dielectric material between plates, as follows:

C = εo A / d, where d is the distance between plates.

If d is reduced to d/2, we can see that it produces as a net effect, to double the capacitance value.

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.