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ELECTRIC FIELD ? Help me in this question.... Thanks ?? ??? plz, refer to attachment ??
question image
Asked by jhajuhi19 | 27 Nov, 2019, 23:50: PM
answered-by-expert Expert Answer
 
Let us consider that origin of x-y coordinate system coincides with the centre of circle as shown in figure
Charge locations are marked as 1,2,3....8 and shown in figure.
 
Charge at [1] and [5] are having equal magnitude and same sign, hence field at centre due to these charges is zero.
 
Charge at [2] and [6] are having equal magnitude but opposite sign, hence field vector at centre is pointing towards point [6]
 
Similarly if we consider other charges which are at line, we get only two electric field vectors F1 and F2 as shown in figure.
 
Resultant electric field vector of F1 and F2 is along negative side of y-axis and its magnitude is √2 F.
( F = |F1| = | F2| )
 
Magnitude of resultant electric field    = (√2 q ) / [ (4πεo) r2 ]
Answered by Thiyagarajan K | 28 Nov, 2019, 09:50: AM

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