Question

Options are:a)4Cnb)5Cnc)6 Cnd)3 Cn​

Answer 1

Answer:

Option B. 5 nC

Explanation:

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

Capicitance (C) = 100 pF

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Next, we shall determine the quantity of charge. This can be obtained as follow:

Capicitance (C) = 1×10¯¹⁰ F

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Q = CV

Q = 1×10¯¹⁰ × 50

Q = 5×10¯⁹ C

Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:

1 C = 1×10⁹ nC

Therefore,

5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C

5×10¯⁹ C = 5 nC

Thus, the quantity of charge is 5 nC