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Using the result of the previous problem, determine the direction and magnitude of the electric field in a plane which is perpendicular to a long, charged rod, and contains one of the rod’s endpoints.

Step-by-step

Using the result of the previous problem, it can be stated that the direction of the electric field at a point on the plane, and a distance h from the end of the infinitely long rod, makes an angle of {{45}^{\omicron }} with the rod.
The magnitude E of the electric field strength can be found using the following ‘trick’. Imagine two very long, uniformly charged rods joined end to end. The resultant field strength will be the vector sum of the field strengths of the two ‘half-rods’.
The direction of the results will obviously be perpendicular to the rod –in view of the symmetry – and its magnitude, {\sqrt {2}} times the field strength E for an individual rod, can be found using Gauss’s law for electric field lines. Enclose a section of length ‘ of the infinitely long rod in a notional cylinder of radius h.
There is a charge Q = ‘σ inside the cylinder, where σ is the charge per unit length, and the number of field lines crossing the cylinder (the electric flux) is {{\phi }_{E}}={\sqrt {2}}E2πhl.
According to Gauss’s law, {{\xi }_{0}{{\phi }_{E}}}=Q,givingE={\frac {\sqrt {2}} {4{\pi }{{\xi }_{0}}}}{\frac {\sigma } {h}}.

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