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The quarter disc of radius R (see figure) has a uniform surface charge density σ. The Z component of electric field at (0,0,Z) is found to be EZ=σfϵ0[1−Z√R2+Z2] Then, the value of 2f is

Answer» The quarter disc of radius R (see figure) has a uniform surface charge density σ.
The Z component of electric field at (0,0,Z) is found to be EZ=σfϵ0[1ZR2+Z2]
Then, the value of 2f is


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