What Have I Learned So Far? 1. Solve for the capacitance of a capacitor if it stores a charge of $$ 1.3 imes 10^{4} \mathrm{C} $$ when delivered with a voltage of $$ 3 imes 10^{3} \mathrm{~V} $$. 2. How much charge can a $$ 4.00-\mu \mathrm{F} $$ capacitor store on its plates when it is connected to a 12.0 - V battery? 3. A capacitor with air between its plates has a capacitance of $$ 5 \mu \mathrm{~F} $$. What is its capacitance when dry wax with a dielectric constant of 2.7 is placed between the plates? Compute for the charge on each plate of a $$ 0.07 \mu \mathrm{~F} $$ capacitor when the potential difference between the plates is 210 V . A capacitor is charged with 8.79 nC and has a $$ 150-\mathrm{V} $$ potential difference between its terminals. What is the capacitance offered by the capacitor? 4.

Question

What Have I Learned So Far? 1. Solve for the capacitance of a capacitor if it stores a charge of $$ 1.3 	imes 10^{4} \mathrm{C} $$ when delivered with a voltage of $$ 3 	imes 10^{3} \mathrm{~V} $$. 2. How much charge can a $$ 4.00-\mu \mathrm{F} $$ capacitor store on its plates when it is connected to a 12.0 - V battery? 3. A capacitor with air between its plates has a capacitance of $$ 5 \mu \mathrm{~F} $$. What is its capacitance when dry wax with a dielectric constant of 2.7 is placed between the plates? Compute for the charge on each plate of a $$ 0.07 \mu \mathrm{~F} $$ capacitor when the potential difference between the plates is 210 V . A capacitor is charged with 8.79 nC and has a $$ 150-\mathrm{V} $$ potential difference between its terminals. What is the capacitance offered by the capacitor? 4.

UpStudy AI · Accepted Answer

Sure! Let's tackle each of these capacitor-related problems step by step.

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### **1. Calculating Capacitance**

**Problem:**  
Solve for the capacitance of a capacitor that stores a charge of $$ 1.3 \times 10^{4} \, \mathrm{C} $$ when a voltage of $$ 3 \times 10^{3} \, \mathrm{V} $$ is applied.

**Solution:**  
The relationship between charge ($$ Q $$), capacitance ($$ C $$), and voltage ($$ V $$) is given by:

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

Plugging in the given values:

$$
C = \frac{1.3 \times 10^{4} \, \mathrm{C}}{3 \times 10^{3} \, \mathrm{V}} = \frac{1.3 \times 10^{4}}{3 \times 10^{3}} \, \mathrm{F} = 4.333... \, \mathrm{F}
$$

**Answer:**  
The capacitance is approximately **4.33 farads (F)**.

---

### **2. Determining Stored Charge**

**Problem:**  
How much charge can a $$ 4.00 \, \mu\mathrm{F} $$ capacitor store on its plates when connected to a 12.0 V battery?

**Solution:**  
Using the formula:

$$
Q = C \times V
$$

Where:
- $$ C = 4.00 \times 10^{-6} \, \mathrm{F} $$
- $$ V = 12.0 \, \mathrm{V} $$

$$
Q = 4.00 \times 10^{-6} \, \mathrm{F} \times 12.0 \, \mathrm{V} = 4.8 \times 10^{-5} \, \mathrm{C}
$$

**Answer:**  
The capacitor stores **4.8 × 10⁻⁵ coulombs (48 µC)** of charge.

---

### **3. Capacitance with a Dielectric**

**Problem:**  
A capacitor with air between its plates has a capacitance of $$ 5 \, \mu\mathrm{F} $$. What is its capacitance when dry wax with a dielectric constant of 2.7 is placed between the plates?

**Solution:**  
The capacitance with a dielectric ($$ C' $$) is related to the original capacitance ($$ C $$) by the dielectric constant ($$ \kappa $$):

$$
C' = \kappa \times C
$$

Given:
- $$ \kappa = 2.7 $$
- $$ C = 5 \times 10^{-6} \, \mathrm{F} $$

$$
C' = 2.7 \times 5 \times 10^{-6} \, \mathrm{F} = 13.5 \times 10^{-6} \, \mathrm{F} = 13.5 \, \mu\mathrm{F}
$$

**Answer:**  
The new capacitance is **13.5 µF**.

---

### **4. Computing Charge on a Capacitor**

**Problem:**  
Compute the charge on each plate of a $$ 0.07 \, \mu\mathrm{F} $$ capacitor when the potential difference between the plates is 210 V.

**Solution:**  
Using the formula:

$$
Q = C \times V
$$

Where:
- $$ C = 0.07 \times 10^{-6} \, \mathrm{F} $$
- $$ V = 210 \, \mathrm{V} $$

$$
Q = 0.07 \times 10^{-6} \, \mathrm{F} \times 210 \, \mathrm{V} = 14.7 \times 10^{-6} \, \mathrm{C} = 1.47 \times 10^{-5} \, \mathrm{C}
$$

**Answer:**  
Each plate holds **1.47 × 10⁻⁵ coulombs (14.7 µC)** of charge.

---

### **5. Determining Capacitance from Charge and Voltage**

**Problem:**  
A capacitor is charged with 8.79 nC and has a 150 V potential difference between its terminals. What is the capacitance offered by the capacitor?

**Solution:**  
Using the formula:

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

Where:
- $$ Q = 8.79 \times 10^{-9} \, \mathrm{C} $$
- $$ V = 150 \, \mathrm{V} $$

$$
C = \frac{8.79 \times 10^{-9} \, \mathrm{C}}{150 \, \mathrm{V}} = 5.86 \times 10^{-11} \, \mathrm{F} = 58.6 \, \mathrm{pF}
$$

**Answer:**  
The capacitance is approximately **58.6 picofarads (pF)**.

---

If you have any further questions or need clarification on any of these problems, feel free to ask!
Here are the simplified answers to the capacitor problems: 1. **Capacitance Calculation:** - **Answer:** 4.33 farads (F) 2. **Stored Charge:** - **Answer:** 4.8 × 10⁻⁵ coulombs (48 µC) 3. **Capacitance with Dielectric:** - **Answer:** 13.5 µF 4. **Charge on Capacitor:** - **Answer:** 1.47 × 10⁻⁵ coulombs (14.7 µC) on each plate 5. **Capacitance from Charge and Voltage:** - **Answer:** 58.6 picofarads (pF)

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