A rod 12.0 cm long is uniformly charged and has a total charge of-20.0 μC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center.
magnitude _______
direction _______
Solution)
Given :
length L = 12 cm = 0.12 m
Charge of rod = -20 x 10^-6 C
Distance of point P from center of rod = 32 cm
We know, E = kQ/(D+L)(D)
where D = distance from end point of rod to P
So, D= 32-6 = 26 cm = 0.26 m
Now, E = (9.0 x 10^9 N m^2 / C^2)(-20 x 10^-6 C)/(0.26+0.12)(0.26) = -1.82x 10^6 V/m
Direction' towards the center of the rod because of -ve charge
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