Answer :
The separation between two plates of parallel capacitor is 8.51 mm and when the separation is doubled the capacitance is reduced to halve.
Magnitude of Charge (Q) = 290 pc
Potential difference between plates (V) = 41.0 V
Let capacitance be = C
Using the relationship between charge and potential difference,
= Q = C × V
= C = Q / V
= C = 290 × 10⁻¹² / 41
= C = 7.07 × 10⁻¹² F
Now, we know that C = (ε × A) / d
Here, Area (A) = 6.80 cm² and
Distance = d
Thus, distance =
= d = (ε × A) / C
= d = (8.85 × 10⁻¹² × 6.80 × 10⁻⁴) / (7.07 × 10⁻¹²)
= d = 8.51 × 10⁻⁴ m
= d = 8.51 mm
Using Q = C × V,
= V = Q/C
If the separation between the two plates is doubled, then the capacitance is halve of the original value.
Thus, new V =
= 41 × 2
= 82 V
Hence, the new capacitance is 82 V when the distance of plates is doubles and the original separation between plates is 8.51 mm.
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