Question

Two conductors having net charges of +16.0 HC and -16.0 HC have a potential difference of 16.0 V between them. (a) Determine the capacitance of the system (b) What is the potential difference between the two conductors if the charges on each are increased to +256.0 C and-256.0 uC

Answer 1

The capacitance of the system is 1.0 F. The potential difference between the two conductors when the charges are increased is +256.0 V.

Step-by-step explanation:

(a) Determine the capacitance of the system:

The potential difference between the two conductors is given as 16.0 V. The net charges on the conductors are +16.0 HC and -16.0 HC.

The capacitance of the system can be determined using the formula:

C = Q / V

where C is the capacitance, Q is the charge on one conductor, and V is the potential difference between the conductors.

Plugging in the values, we get:

C = 16.0 HC / 16.0 V = 1.0 F

Therefore, the capacitance of the system is 1.0 F.

(b) What is the potential difference between the two conductors if the charges on each are increased to +256.0 C and-256.0 uC:

The capacitance of the system does not change when the charges on the conductors are increased. Using the formula:

V = Q / C

we can calculate the new potential difference, where Q is the new charge on one conductor and C is the capacitance of the system.

Plugging in the values, we get:

V = +256.0 C / 1.0 F = +256.0 V

Therefore, the potential difference between the two conductors when the charges are increased is +256.0 V.

Answer 2

(a) Capacitance is
`1\mu C`

(b) Potential difference is 256 V

Solution:

As per the question:

Charges on the two conductors, q is
`+ 16\muC` and
`- 16\muC`

The potential difference between the charges, V = 16.0 V

Now,

(a) The capacitance of the system is given by:

q = CV

`C = (q)/(V) = (16* 10^(- 6))/(16) = 1\mu F`

(Since,
`1\mu C = 10^(- 6) C`)

Now,

(b) When the charge, 'q' is increased to
`+ 256\mu C` and
`- 256\mu C`

q = CV

`V = (q)/(C) = (256* 10^(- 6))/(1* 10^(- 6)) = 256\ V`