The quantity of charge q (in coulombs) that has passed through a surface of area 2.10 cm² varies with time according to the equation q = 4t³ + 5t + 6, where t is in seconds. (a) What is the instantaneous current through the surface at t = 0.900 s? A (b) What is the value of the current density? kA/m²
The quantity of charge q (in coulombs) that has passed through a surface of area 2.10 cm² varies with time according to the equation q = 4t³ + 5t + 6, where t is in seconds. (a) What is the instantaneous current through the surface at t = 0.900 s? A (b) What is the value of the current density? kA/m²
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![**Electrical Engineering Problem Solving**
The quantity of charge \( q \) (in coulombs) that has passed through a surface of area \( 2.10 \, \text{cm}^2 \) varies with time according to the equation \( q = 4t^3 + 5t + 6 \), where \( t \) is in seconds.
**(a) What is the instantaneous current through the surface at \( t = 0.900 \, \text{s} \)?**
\[ I = \, \_\_\_\_ \, \text{A} \]
**(b) What is the value of the current density?**
\[ J = \, \_\_\_\_ \, \text{kA/m}^2 \]
**Explanation:**
1. To determine the instantaneous current \( I(t) \) at any time \( t \), we take the derivative of \( q \) with respect to \( t \):
\[ I(t) = \frac{dq}{dt} \]
2. The current density \( J \) can be found using the formula:
\[ J = \frac{I}{A} \]
where \( A \) is the area in square meters, converted from \( 2.10 \, \text{cm}^2 \).
This problem involves derivatives and practical application of electric current and current density concepts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F800a48ac-05a9-41c6-9b96-1b31e1f3fea8%2Fe8a6bfbf-132d-4c30-8ab5-b063f720216a%2Ff2mgxq_processed.png&w=3840&q=75)
Transcribed Image Text:**Electrical Engineering Problem Solving**
The quantity of charge \( q \) (in coulombs) that has passed through a surface of area \( 2.10 \, \text{cm}^2 \) varies with time according to the equation \( q = 4t^3 + 5t + 6 \), where \( t \) is in seconds.
**(a) What is the instantaneous current through the surface at \( t = 0.900 \, \text{s} \)?**
\[ I = \, \_\_\_\_ \, \text{A} \]
**(b) What is the value of the current density?**
\[ J = \, \_\_\_\_ \, \text{kA/m}^2 \]
**Explanation:**
1. To determine the instantaneous current \( I(t) \) at any time \( t \), we take the derivative of \( q \) with respect to \( t \):
\[ I(t) = \frac{dq}{dt} \]
2. The current density \( J \) can be found using the formula:
\[ J = \frac{I}{A} \]
where \( A \) is the area in square meters, converted from \( 2.10 \, \text{cm}^2 \).
This problem involves derivatives and practical application of electric current and current density concepts.
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