Answer:
V₂ = 700 V
Explanation:
- By definition, the capacitance of a capacitor is given by the following expression:
- where Q is the charge on one of the plates and V is the potential difference between them.
- For a parallel-plate capacitor, it can be showed (just applying Gauss' Law to the surface of one of the plates), that the capacitance can be written as follows:
- where ε₀ = 8.85*10⁻¹² F/m, A is the area of one of the plates and d is the distance between the plates.
- When the plates are pulled to a separation that is five times the original one, we can see that the new value of the capacitance, is 5 times less than the original one, as the distance d between plates is in the denominator.
- As the charge can't change, because the capacitor is isolated, if the new value is 5 times smaller, the potential difference must be 5 times larger.
- V₂ = V₁* 5 = 140 V * 5 = 700 V