Q. 41

# A 10 cm long rod

Given:
Length of the rod : L = 10cm = 0.1m
Charge on the rod; q = +50 μC= 50× 10-6 C Here, C is 10 cm = 0.1 m away from both ends of the rod AB.
Distance between C and centre of AB : r
Formula used:
From Pythagoras Theorem,
r2+(L/2)2=(BC)2

r2= [0.1]2- [0.05]2

r2= 7.5×10-3

r=√ (7.5×10-3)

r=0.0866 m
Now we know that, Electric field at a point on the perpendicular bisector of a uniformly charged rod is: Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge, L is the length of the rod and Q is the magnitude of the charge.
Substituting we get,  Hence, electric field at a point 10 cm away from the ends of the rod is 5.2 × 107 NC-1.

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