dc electric circuits

a capacitor has plates with an area of $1.64 \times 10^{-3}\text{ m}^2$.

how far apart should the plates be to create a capacitance of $2.38 \times 10^{-9}\text{ f}$?

$? \times 10^{?}\text{ m}$

$\varepsilon_0 = 8.85 \cdot 10^{-12}\text{ c}^2/\text{n} \cdot \text{m}^2$

Question

dc electric circuits

a capacitor has plates with an area of $1.64 \times 10^{-3}\text{ m}^2$.

how far apart should the plates be to create a capacitance of $2.38 \times 10^{-9}\text{ f}$?

$? \times 10^{?}\text{ m}$

$\varepsilon_0 = 8.85 \cdot 10^{-12}\text{ c}^2/\text{n} \cdot \text{m}^2$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Parallel Plate Capacitor" ], "new_concepts": [], "current_concepts": [ "Capacitance", "Parallel Plate Capacitor" ] } ## Identify the given values Using the *Parallel Plate Capacitor* knowledge point $$ \begin{aligned} A &= 1.64 \times 10^{-3}\text{ m}^2 \ C &= 2.38 \times 10^{-9}\text{ F} \ \varepsilon_0 &= 8.85 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2) \end{aligned} $$ ## Solve for the plate separation Using the *Parallel Plate Capacitor* knowledge point $$ \begin{aligned} C &= \frac{\varepsilon_0 A}{d} \ d &= \frac{\varepsilon_0 A}{C} \ d &= \frac{(8.85 \times 10^{-12})(1.64 \times 10^{-3})}{2.38 \times 10^{-9}} \ d &= \frac{1.4514 \times 10^{-14}}{2.38 \times 10^{-9}} \ d &\approx 6.10 \times 10^{-6}\text{ m} \end{aligned} $$ A capacitor has plates with an area of $1.64 \times 10^{-3}\text{ m}^2$. How far apart should the plates be to create a capacitance of $2.38 \times 10^{-9}\text{ F}$? $6.10 \times 10^{-6}$ m { "subject": "Natural Science", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Natural Science", "Physics", "Parallel Plate Capacitor" ] }

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Question

Explanation:

Identify the given values

Solve for the plate separation

Answer: