Question

The two metal objects in Figure have net charges of +70pC and -7pC , which result in a 20 Vpotential difference between them. (a) What is the capacitance of the system? (b) If the charges are changed to+200 pCand -200 pCwhat does the capacitance become? (c) What does the potential difference become?

Short answer

1. The capacitance of the system is C= 3.5 pF .
2. The capacitance remains the same as C = 3.5 pF .
3. The potential difference is 57 V

Step-by-step solution

Given data:

The two metal objects have net charge of q = +70 pC and q = -70 pC .

The potential difference is V = 20 V .

The charges are charged to +200 pC and -200 pC .

Determining the concept:

Define the capacitance of the system at the given charges. Also, find capacitance if the charges are charged to +200 pCand -200 pC. Then the potential difference can be calculated using the same formula.

Formula:

Write the equation for capacitance as below.

$C=\frac{q}{v}$ ….. (1)

Where, C is the capacitance, V is the potential difference, and q is the charge on the particle.

(a) Determining the capacitance of the system:

$$\begin{gathered} \mathrm{C}=\frac{70\mathrm{pC}}{20\mathrm{V}} \\ =3.5\mathrm{pF} \end{gathered}$$From equation (1), the capacitance of the system is given by,

$C=\frac{q}{v}$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{C}=\frac{70\mathrm{pC}}{20\mathrm{V}} \\ =3.5\mathrm{pF} \end{gathered}$$

Hence, the capacitance of the system is 3.5 pF .

(b) Determining the change in capacitance:

The capacitance depends on the geometry of objects and not on the charge on it.

Hence, the capacitance remains the same. That is,

C = 3.5 pF

(c) Determining the difference in potential:

From equation (1), the potential difference is given by,

$V=\frac{q}{c}$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{V}=\frac{200\mathrm{pC}}{3.5\mathrm{pF}} \\ =57.143\mathrm{V} \\ \approx \end{gathered}$$

Hence, the potential difference is 57 V .