Calculate the amount of potential difference in a capacitor of 0.9 MF , If the amount of charge is 1.4x10^-4 C .(Show all the work)

1.56×10¯¹⁰ V.

Explanation:

From the question given above, the following data were obtained:

Capicitance (C) = 0.9 MF

Quantity of electricity (Q) = 1.4x10¯⁴ C

Potential difference (V) =?

Next, we shall convert 0.9 MF to farad (F). This can be obtained as follow:

1 MF = 10⁶ F

Therefore,

0.9 MF = 0.9 MF × 10⁶ F / 1 MF

0.9 MF = 9×10⁵ F

Finally, we shall determine the potential difference. This can be obtained as follow:

Capicitance (C) = 9×10⁵ F

Quantity of electricity (Q) = 1.4x10¯⁴ C

Potential difference (V) =?

Q = CV

1.4x10¯⁴ = 9×10⁵ × V

Divide both side by 9×10⁵

V = 1.4x10¯⁴ / 9×10⁵

V = 1.56×10¯¹⁰ V

Therefore, the potential difference is 1.56×10¯¹⁰ V.

the first, second, and fifth options

2b2t

explanation:

2b2t