Electric field outside a long wire carrying charge q is proportional to :

Q: Electric field outside a long wire carrying charge q is proportional to :

Using Gauss Law across the cylindrical Gaussian surface,\$\displaystyle\int E.dS = \dfrac{q_{enc}}{\epsilon_0}\$\$\therefore E\times 2\pi r \times l = \dfrac{\lambda\times l}{\epsilon_0}\$\$\therefore E = \dfrac{\lambda}{2\pi r\epsilon_0}\$Here \$\lambda\$ is the linear charge density on the wire.

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Using Gauss Law across the cylindrical Gaussian surface,$\displaystyle\int E.dS = \dfrac{q_{enc}}{\epsilon_0}$$\therefore E\times 2\pi r \times l = \dfrac{\lambda\times l}{\epsilon_0}$$\therefore E = \dfrac{\lambda}{2\pi r\epsilon_0}$Here $\lambda$ is the linear charge density on the wire.

Electric Charges And Fields

Physics
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