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} \

Question

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} \

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": [ "Parallel Plate Capacitor", "Capacitance" ] } ## Identify given parameters Using the *Parallel Plate Capacitor* knowledge point $$ \begin{aligned} A &= 1.64 \times 10^{-3}\text{ m}^2\ C &= 2.38 \times 10^{-9}\text{ F}\ \varepsilon_0 &\approx 8.85 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2) \end{aligned} $$ ## Solve for distance Using the *Parallel Plate Capacitor* knowledge point $$ \begin{aligned} C &= \frac{\varepsilon_0 A}{d}\ d &= \frac{\varepsilon_0 A}{C}\ d &= \frac{(8.85 \times 10^{-12}\text{ F/m})(1.64 \times 10^{-3}\text{ m}^2)}{2.38 \times 10^{-9}\text{ F}}\ d &\approx 6.10 \times 10^{-6}\text{ m} \end{aligned} $$ 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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Identify given parameters

Solve for distance

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