V = IR
I = current
R = resistance
Voltage = 100 * (3.44x 10^-4) = do the calculation
Hope this helps
Answer:
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.
