Answer :
- Use the formula for capacitance: $C = \frac{{\epsilon_0 A}}{{d}}$.
- Rearrange the formula to solve for $d$: $d = \frac{{\epsilon_0 A}}{{C}}$.
- Substitute the given values: $d = \frac{{(8.85 \times 10^{-12})(1.64 \times 10^{-3})}}{{2.38 \times 10^{-9}}}$.
- Calculate the distance: $d = 6.098319327731092 \times 10^{-6} m \approx 6.10 \times 10^{-6} m$.
- The final answer is $\boxed{6.10 \times 10^{-6} m}$.
### Explanation
1. Understanding the Problem
We are given the area of the capacitor plates $A = 1.64 Imes 10^{-3} m^2$, the capacitance $C = 2.38 Imes 10^{-9} F$, and the permittivity of free space $\epsilon_0 = 8.85 Imes 10^{-12} C^2 / N Imes m^2$. We need to find the distance $d$ between the plates. The formula for capacitance is given by $C = \frac{\epsilon_0 A}{d}$.
2. Rearranging the Formula
We need to rearrange the formula to solve for $d$. Multiplying both sides by $d$ and dividing by $C$, we get:$$d = \frac{\epsilon_0 A}{C}$$
3. Substituting the Values
Now, we substitute the given values into the formula:$$d = \frac{(8.85 Imes 10^{-12} C^2 / N Imes m^2) Imes (1.64 Imes 10^{-3} m^2)}{2.38 Imes 10^{-9} F}$$
4. Calculating the Distance
Calculating the value of $d$:$$d = \frac{8.85 Imes 1.64 Imes 10^{-15}}{2.38 Imes 10^{-9}} = \frac{14.514 Imes 10^{-15}}{2.38 Imes 10^{-9}} = 6.098319327731092 Imes 10^{-6} m$$
5. Expressing the Answer
We need to express the answer in the form $[?] Imes 10^{?} m$. From the calculation above, we have $d = 6.098319327731092 Imes 10^{-6} m$. Therefore, the coefficient is approximately 6.10 and the exponent is -6.
6. Final Answer
The distance between the plates should be approximately $6.10 Imes 10^{-6} m$.
### Examples
Capacitors are essential components in electronic devices, storing electrical energy. This calculation is crucial in designing capacitors for specific applications, such as in smartphones, computers, and power supplies. By adjusting the distance between the plates, engineers can fine-tune the capacitance to meet the circuit's requirements, ensuring optimal performance and efficiency. Understanding the relationship between plate distance and capacitance allows for the creation of smaller, more powerful electronic devices.
- Rearrange the formula to solve for $d$: $d = \frac{{\epsilon_0 A}}{{C}}$.
- Substitute the given values: $d = \frac{{(8.85 \times 10^{-12})(1.64 \times 10^{-3})}}{{2.38 \times 10^{-9}}}$.
- Calculate the distance: $d = 6.098319327731092 \times 10^{-6} m \approx 6.10 \times 10^{-6} m$.
- The final answer is $\boxed{6.10 \times 10^{-6} m}$.
### Explanation
1. Understanding the Problem
We are given the area of the capacitor plates $A = 1.64 Imes 10^{-3} m^2$, the capacitance $C = 2.38 Imes 10^{-9} F$, and the permittivity of free space $\epsilon_0 = 8.85 Imes 10^{-12} C^2 / N Imes m^2$. We need to find the distance $d$ between the plates. The formula for capacitance is given by $C = \frac{\epsilon_0 A}{d}$.
2. Rearranging the Formula
We need to rearrange the formula to solve for $d$. Multiplying both sides by $d$ and dividing by $C$, we get:$$d = \frac{\epsilon_0 A}{C}$$
3. Substituting the Values
Now, we substitute the given values into the formula:$$d = \frac{(8.85 Imes 10^{-12} C^2 / N Imes m^2) Imes (1.64 Imes 10^{-3} m^2)}{2.38 Imes 10^{-9} F}$$
4. Calculating the Distance
Calculating the value of $d$:$$d = \frac{8.85 Imes 1.64 Imes 10^{-15}}{2.38 Imes 10^{-9}} = \frac{14.514 Imes 10^{-15}}{2.38 Imes 10^{-9}} = 6.098319327731092 Imes 10^{-6} m$$
5. Expressing the Answer
We need to express the answer in the form $[?] Imes 10^{?} m$. From the calculation above, we have $d = 6.098319327731092 Imes 10^{-6} m$. Therefore, the coefficient is approximately 6.10 and the exponent is -6.
6. Final Answer
The distance between the plates should be approximately $6.10 Imes 10^{-6} m$.
### Examples
Capacitors are essential components in electronic devices, storing electrical energy. This calculation is crucial in designing capacitors for specific applications, such as in smartphones, computers, and power supplies. By adjusting the distance between the plates, engineers can fine-tune the capacitance to meet the circuit's requirements, ensuring optimal performance and efficiency. Understanding the relationship between plate distance and capacitance allows for the creation of smaller, more powerful electronic devices.
