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.
I think you can google this because I really don’t know the answer I’m so sorry
The answer is D. Products are formed from reactants by the breaking and forming of new bonds.
Answer:
12500 V
Explanation:
The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:
where
is the potential difference between the plates
E is the electric field strength
d is the distance between the plates
For the capacitor in this problem, we have
Substituting, we find
The answer is c: <span>1960 J
</span>Potential Energy :
<span>PE = m x g x h = 40*9.8*5=1960
</span>