Answer:
1.0 volts
Explanation:
Given that:
The potential difference between the plate V = 10 V
C = 0.160 pF = 0.160 × 10⁻¹² F
The charge on the capacitor
Q = CV
Q = 0.160 × 10⁻¹² × 10
Q = 1.6 × 10⁻¹² C
It implies that the positive plate of capacitor has +1.6 × 10⁻¹² C charge while the negative plate has -1.6 × 10⁻¹² C
The number of excess electrons on the negative plate is:
[tex]n = \dfrac{q}{e}[/tex]
[tex]n = \dfrac{-1.6\times 10^{-12}}{-1.6\times 10^{-19}}[/tex]
[tex]n = 1.0 \times 10^7[/tex]
Thus, electron deficiency on the positive plate is [tex]1.0 \times 10^7[/tex]
The net negative charge that moves towards the positive plate is :
q = number of electrons moved × e
[tex]q = 1\times 10^6 \times (-1.6 \times 10^{-19})[/tex]
[tex]q = -1.6 \times 10^{-13} \ C[/tex]
Now, the net charge on the positive plate is:
[tex]q_{net} = q +q' \\ \\ q_{net} = (1.6 \times 10^{-12}) + (-1.6 \times 10^{-13})[/tex]
[tex]q_{net} =1.44 \times 10^{-12} \ C[/tex]
The potential difference between the plate;
[tex]V_{new} = \dfrac{q_{net} }{c}[/tex]
[tex]V_{new} = \dfrac{1.44 \times 10^{-12} }{0.16 \times 10^{-12}}[/tex]
[tex]V_{new} = 9.0 V[/tex]
The reduction in potential difference
[tex]\Delta V = V - V_{new}[/tex]
[tex]\Delta V = 10 - 9.0[/tex]
[tex]\mathbf{\Delta V = 1.0 \ volts}[/tex]
