Question

A rod 14.0 cm long is uniformly charged and has a total charge of 222.0 mc. determine (a) the magnitude and (b) the direction of the electric field along the axis of the rod at a point 36.0 cm from its center

Answer 1

Electric field due to a charged rod along its axis is given by

$E = \frac{kQ}{(L+r)(r)}$

here we know that

L = 14 cm

r = distance from end of rod

r = 36 - 7 = 29 cm

Q = 222 mC

now we will have

$E = \frac{(9 \times 10^9)(222 \times 10^{-3})}{(0.29)(0.43)}$

$E = 1.6 \times 10^{10} N/C$