Question

A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0 cmcm from the rod's center. The bead is repelled from the rod with a force of 840 μNμN.What is the total charge on the rod?

Answer 1

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

$E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

$F=qE$

Put the value into the formula

$F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

We need to calculate the total charge on the rod

$Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}$

Put the value into the formula

$Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}$

$Q=47.8\times10^{-9}\ C$

$Q=47.8\ nC$

Hence, The total charge on the rod is 47.8 nC.