Answer:
The potential difference between the plates is 596.2 volts.
Explanation:
Given that,
Capacitance 
Charge 
Separation of plates = 0.313 mm
We need to calculate the potential difference between the plates
Using formula of potential difference
Where, Q = charge
C = capacitance
Put the value into the formula
Hence,The potential difference between the plates is 596.2 volts.
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We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.
Voltage = current(I) * resistance (R)
V = 18
R = 6
18 = I * 6
I = 18/6 = 3 Amps or D
