Acapacitor with plates separated by a distance d is charged to a potential difference δvc. all wires and batteries are disconnected, then the two plates are pulled further apart to a new separation distance 2d. what happens to the potential difference across the capacitor, δv, and the charge on the capacitor, q, as a result of this change?

Explanation:

Initial separation of plate = d

final separation = 2d

The capacitance of the capacitor will reduce from C to C/2 because

capacitance = ε A / d

d is distance between plates.

As the batteries are disconnected , charge on the capacitor becomes fixed .

Initial charge on the capacitor

= Capacitance x potential difference

Q = C ΔV

Final charge will remain unchanged

Final charge = C ΔV

Final capacitance = C/2

Final potential difference = charge / capacitance

= C ΔV /  C/2

= 2 ΔV

Potential difference is doubled after the pates are further separated.

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles