# A metal disc of radius a rotates with a constant angular velocity ω about its axis. The potential difference between…..

Q. A metal disc of radius a rotates with a constant angular velocity ω about its axis. The potential difference between the center and the rim of the disc is (m = mass of electron, e = charge on electron)

(a) $\displaystyle \frac{m\omega^2 a^2}{e}$

(b) $\displaystyle \frac {1}{2} \frac{m\omega^2 a^2}{e}$

(c) $\displaystyle \frac{e\omega^2 a^2}{2m}$

(d) $\displaystyle \frac{e\omega^2 a^2}{m}$

Ans: (b)

Sol:As metal disc rotates , the free electron also rotates with same angular velocity  having acceleration = ω2r . Hence Charged particle must have a force .But charged particle is only influenced by either electric field or magnetic field . Here there is no external magnetic field in question , hence it is electric field .

e E = m ω2r ; Where E = Electric field

$\displaystyle E = \frac{m \omega^2 r}{e}$

Potential difference between center & rim $\displaystyle = \int_{0}^{a} \vec{E}.\vec{dr}$

$\displaystyle = \int_{0}^{a}\frac{m \omega^2 r}{e} dr$

$\displaystyle = [\frac{m \omega^2 r^2}{2 e}]_{0}^{a}$

$\displaystyle = \frac{m \omega^2 a^2}{2 e}$