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Five charges , q each are placed at the corners of a regular pentagon of side. (a) (i) What will be the electric field at 0, the centre of the pentagon? (ii) What will be the electric field at 0 if the charge from one of the corners (say A) is removed? (iii). What will be the electric field at 0 if the pentagon is replaced by n-sided regular polygon with charge q at each of its corner? |
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Answer» Solution :(a) (i) The point O is equidistant from all the charges at the end point of pentagon. THUS, due to symmetry, the forces due to all the charges are cancelled out. As a RESULT ELECTRIC field at O is zero. (ii). When charge q is removed a negative charge will develop at A giving electric field `E=(qxx1)/(4piepsi_(0)r^(2))` along OA. (iii) If charge q at A is replaced by -q, then two negative charges -2q will develop there. thus, the value of electric field `E=(2q)/(4piepsi_(0)r^(2))` along OA. (b). When pentagon is replaced by N sided regular polygon with charge q at each of its corners, the electric field at O would continue to be zero as symmetricity of the charges is dueto the regularity of the polygon it doesn't depend on the number of sided or the number of charges. |
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