Question
A conducting disc of radius $$R$$ rotating about its axis with an angular velocity $$\omega .$$ Then the potential difference between the centre of the disc and its edge is (no magnetic field is present)
A.
zero
B.
$$\frac{{{m_e}{\omega ^2}{R^2}}}{{2e}}$$
C.
$$\frac{{{m_e}\omega {R^3}}}{{3e}}$$
D.
$$\frac{{e{m_e}\omega {R^2}}}{2}$$
Answer :
$$\frac{{{m_e}{\omega ^2}{R^2}}}{{2e}}$$
Solution :
$$\eqalign{ & eE = {m_e}{\omega ^2}r \cr & \Rightarrow \int E \,dr = \frac{{{m_e}{\omega ^2}}}{e}\int\limits_0^R {rdr} \cr & \Rightarrow V = \frac{{{m_e}{\omega ^2}{R^2}}}{{2e}} \cr} $$
$$\eqalign{ & eE = {m_e}{\omega ^2}r \cr & \Rightarrow \int E \,dr = \frac{{{m_e}{\omega ^2}}}{e}\int\limits_0^R {rdr} \cr & \Rightarrow V = \frac{{{m_e}{\omega ^2}{R^2}}}{{2e}} \cr} $$