8. Solve the following linear equation q(t) with conditions. dq :-- IR = 0 (I = , t= 0, q(0) = 0, qo = Cɛ) C dt
8. Solve the following linear equation q(t) with conditions. dq :-- IR = 0 (I = , t= 0, q(0) = 0, qo = Cɛ) C dt
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![### Problem 8: Linear Differential Equation in Physics
#### Task:
**Solve the following linear equation \( q(t) \) with given conditions:**
\[
\varepsilon - \frac{q}{C} - IR = 0 \quad \left(I = \frac{dq}{dt}, \, t = 0, \, q(0) = 0, \, q_0 = C\varepsilon \right)
\]
#### Explanation:
We need to solve the linear differential equation that describes the behavior of charge \( q(t) \) in an electrical circuit involving a capacitor \( C \), resistor \( R \), and an applied voltage \( \varepsilon \). The current \( I \) in the circuit is given by the time derivative of the charge \( I = \frac{dq}{dt} \).
#### Initial and Boundary Conditions:
- At time \( t = 0 \), the charge \( q \) on the capacitor is zero: \( q(0) = 0 \).
- The initial charge \( q_0 \) relates to the product of the capacitance \( C \) and the applied voltage \( \varepsilon \): \( q_0 = C\varepsilon \).
This problem will involve solving the differential equation and applying the initial conditions to find the specific solution for \( q(t) \).
#### Steps to Solve:
1. Substitute \( I = \frac{dq}{dt} \) into the differential equation.
2. Rearrange and simplify the resulting differential equation.
3. Integrate the equation to find \( q(t) \).
4. Apply the initial conditions to determine any constants of integration, obtaining the specific solution.
#### Final Solution:
After working through these steps, you'll have the function \( q(t) \) that describes the charge on the capacitor as a function of time with the applied initial conditions.
Note: This is a classic problem in the study of RC circuits in electrical engineering and physics, illustrating the principles of charging and discharging of capacitors through resistive elements.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e888a60-a63c-44c5-bf2a-af4d3002d5ca%2Fe68a9fdc-ffdb-486c-846c-a7a7cec03c3a%2Fx8swdlh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 8: Linear Differential Equation in Physics
#### Task:
**Solve the following linear equation \( q(t) \) with given conditions:**
\[
\varepsilon - \frac{q}{C} - IR = 0 \quad \left(I = \frac{dq}{dt}, \, t = 0, \, q(0) = 0, \, q_0 = C\varepsilon \right)
\]
#### Explanation:
We need to solve the linear differential equation that describes the behavior of charge \( q(t) \) in an electrical circuit involving a capacitor \( C \), resistor \( R \), and an applied voltage \( \varepsilon \). The current \( I \) in the circuit is given by the time derivative of the charge \( I = \frac{dq}{dt} \).
#### Initial and Boundary Conditions:
- At time \( t = 0 \), the charge \( q \) on the capacitor is zero: \( q(0) = 0 \).
- The initial charge \( q_0 \) relates to the product of the capacitance \( C \) and the applied voltage \( \varepsilon \): \( q_0 = C\varepsilon \).
This problem will involve solving the differential equation and applying the initial conditions to find the specific solution for \( q(t) \).
#### Steps to Solve:
1. Substitute \( I = \frac{dq}{dt} \) into the differential equation.
2. Rearrange and simplify the resulting differential equation.
3. Integrate the equation to find \( q(t) \).
4. Apply the initial conditions to determine any constants of integration, obtaining the specific solution.
#### Final Solution:
After working through these steps, you'll have the function \( q(t) \) that describes the charge on the capacitor as a function of time with the applied initial conditions.
Note: This is a classic problem in the study of RC circuits in electrical engineering and physics, illustrating the principles of charging and discharging of capacitors through resistive elements.
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