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A circular copper ring of radius r, placed in vacuum, has a charge q on it. Find out the electric fields For what value of x would the electric field be maximum? |
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Answer» Solution :For E to be MAXIMUM, `(DE)/(dx)=0` Now, `(dE)/(dx)=q/(4piepsilon_(0))cdotd/(dx)[x(R^(2)+x^(2))^(-3//2)]` `=q/(4piepsilon_(0))[1cdot(r^(2)+x^(2))^(-3//2)+xcdot(-3/2)cdot(r^(2)+x^(2)^(-5//2)cdot2x]` `=q/(4piepsilon_(0))(r^(2)+x^(2))^(-5//2)[(r^(2)+x^(2))-3x^(2)]` `=q/(4piepsilon_(0))(r^(2)+x^(2))^(-5//2)(r^(2)-2x^(2))` For maximum E, `(dE)/(dx)=0` or `r^(2)-2x^(2)=0` or, `x=pmr/(sqrt2)` The `'pm'` sign implies that the electric FIELD on the axis will be maximum at a distance `r/(sqrt2)` on either side of the RING. |
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