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2 marks) Use Gauss’ law to find the E-field outside of a long straight wire.

Question

2 marks) Use Gauss’ law to find the E-field outside of a long straight wire.

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Solution

Sure, here is the step-by-step solution:

Step 1: Gauss's law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The Gaussian surface we choose is a cylindrical surface with the wire as the axis.

Step 2: The electric field E is radially outward and is constant on the cylindrical surface due to symmetry. Therefore, the electric flux through the curved surface of the cylinder is E * (2πrl), where r is the radius of the cylinder and l is the length.

Step 3: According to Gauss's law, this must be equal to the charge enclosed divided by the permittivity. The charge enclosed is λl, where λ is the linear charge density.

Step 4: Equating the two expressions gives E * (2πrl) = λl / ε₀. Simplifying this gives E = λ / (2πε₀r).

So, the electric field E outside a long straight wire of linear charge density λ is given by E = λ / (2πε₀r).

This problem has been solved

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