Chapter 25: Q2P (page 739)

The capacitor in Fig. 25-25 has a capacitance of $25 μF$ and is initially uncharged. The battery provides a potential difference of 120 VAfter switchis closed, how much charge will pass through it?

Short Answer

Expert verified

Charge passing through the battery is $q = 3.0 \times 10^{- 3} C$

Step by step solution

Given

The capacitance of the capacitor is $C = 25 μF (\frac{10^{- 6} F}{1 μF}) = 2.5 \times 10^{5} F$ .

The potential difference is V = 120 V.

Determining the concept

Using equation 25-1, find the amount of charge that passes through the battery.

Formulae are as follows:

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

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

(a) Determining the charge passing through the battery

From equation 25-1 we know that,

q = CV

The charge passing through the battery is,

$q = (2.5 \times 10^{- 5} F) (120 V) = 3.0 \times 10^{- 3} C$

Hence, the charge passing through the battery is $3.0 \times 10^{- 3} C$

Therefore, by using the formula for capacitance find the charge passing through the battery.

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The capacitor in Fig. 25-25 has a capacitance of $25 μF$ and is initially uncharged. The battery provides a potential difference of 120 VAfter switchis closed, how much charge will pass through it?

Short Answer

Step by step solution

Given

Determining the concept

(a) Determining the charge passing through the battery

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The capacitor in Fig. 25-25 has a capacitance of25μF and is initially uncharged. The battery provides a potential difference of 120 VAfter switchis closed, how much charge will pass through it?

Short Answer

Step by step solution

Given

Determining the concept

(a) Determining the charge passing through the battery

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Famous Physicists

Kinematics Physics

Magnetism

Atoms and Radioactivity

Wave Optics

Study anywhere. Anytime. Across all devices.

The capacitor in Fig. 25-25 has a capacitance of $25 μF$ and is initially uncharged. The battery provides a potential difference of 120 VAfter switchis closed, how much charge will pass through it?