PRACTICE EXERCISE 6.1 In a one-dimensional device, the charge density is given by p、 = p。x/a. If E = 0 at x = 0 and V = 0 at x = a, find V and E. Answer: Po σεα Pot² - (a³ – x³), ² - 2aɛ ax.

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**PRACTICE EXERCISE 6.1**

In a one-dimensional device, the charge density is given by \( \rho_v = \rho_o x / a \). If \( E = 0 \) at \( x = 0 \) and \( V = 0 \) at \( x = a \), find \( V \) and \( E \).

**Answer:**

\[
\frac{\rho_o}{6 \varepsilon a} \left( a^3 - x^3 \right), \frac{\rho_o x^2}{2 a \varepsilon} a_x
\]

*Explanation*: This equation provides expressions to calculate the potential \( V \) and electric field \( E \) based on the given charge density function and boundary conditions. Here, \( \rho_o \) is a constant related to charge density, \( \varepsilon \) is the permittivity of the medium, and \( a \) is a constant that defines limits for \( x \).
Transcribed Image Text:**PRACTICE EXERCISE 6.1** In a one-dimensional device, the charge density is given by \( \rho_v = \rho_o x / a \). If \( E = 0 \) at \( x = 0 \) and \( V = 0 \) at \( x = a \), find \( V \) and \( E \). **Answer:** \[ \frac{\rho_o}{6 \varepsilon a} \left( a^3 - x^3 \right), \frac{\rho_o x^2}{2 a \varepsilon} a_x \] *Explanation*: This equation provides expressions to calculate the potential \( V \) and electric field \( E \) based on the given charge density function and boundary conditions. Here, \( \rho_o \) is a constant related to charge density, \( \varepsilon \) is the permittivity of the medium, and \( a \) is a constant that defines limits for \( x \).
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