Example – 1 Charges with magnitude q are placed at 4 corners of a regular pentagon. These charges
are at distance ‘a’ from the centre of the pentagon. Find electric field intensity at the
centre of the pentagon.
Solution : Charges are placed at corners A, B, C and D of the pentagon. If charge q is placed
at the fifth corner also then by symmetry the intensity ®E at centre O is zero.Let ®E1 be the resultant electric field due to charge at A, B C and D and let ®E2 be the elecric field due to the fifth charge.
Now, ®E1 + ®E2 = 0
\ ®E1 = -®E2
\ Electric field due to charges at A, B, C and D.
Example – 2 Three charges +q, +q, +2q are arranged as shown in figure. What is the field at point P (center of side AC)
Solution : The sum of fields at P due to charges at A and C add up to zero (because of equal magnitude and opposite direction) . Thus the net field at P is that due to +2q charge. Its direction is along the line BP and its magnitude is
Example – 3 An infinite plane of positive charge has a surface charge density s A metal ball B of mass m and charge q is attached to a thread and tied to a point A on the sheet PQ. Find the angle q which AB makes with the plane PQ.
Solution : Due to positive charge the ball will experience electrical force Fe = qE horizontally away from the sheet while the weight of the ball will act vertically downwards and hence if
T is the tension in the string, for equilibrium of ball :
Along horizontal, Tsinq = qE
And along vertical, Tcosq = mg
So , tanq = (qE/mg)
and T = [(mg)2 + (qE)2]1/2
The field E produced by the sheet of charge PQ having charge density s is E = (s/2e0)
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