Question

A rod 10.0 cm long is uniformly charged and has a total charge of -21.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 34.0 cm from its center.

Answer 1

Answer:

The magnitude of electric field is $2.3\times10^6\ C/m^2$ and the direction is inward.

Explanation:

Given that,

Length of rod = 10.0 cm

Charge$Q = -21.0\times10^{-6}\ C$

Distance D = (34.0-10) cm = 0.24 m

We need to calculate the electric field along the axis

$E=\dfrac{kQ}{D(D+L)}$

Where, Q = charge

D = distance

Put the value into the formula

$E = \dfrac{9\times10^{9}\times(-21\times10^{-6}}{0.24(0.24+0.10)}$

$E=-2.3\times10^6\ C/m^2$

Negative sign shows the direction of electric field is inward that means toward the center.

Hence, The magnitude of electric field is $2.3\times10^6\ C/m^2$ and the direction is inward.