A disc having uniform surface charge density $+\sigma$  in upper half and $-\sigma$ in lower half is placed on a rough horizontal surface as shown in figure. A uniform electric field is set up as shown. If mass of disc is M and its radius is R, as well sufficient friction is present to prevent slipping, the acceleration of disc is

[Step 1]: Dipole moment

  $\begin{array}{l}\left(\begin{array}{l}\because \sigma =\frac{q}{A}\\ \Rightarrow q=\sigma A_{hemisphere}=\frac{\sigma \pi R^2}{2}\end{array}\right)\text{}\\ \Rightarrow P=2aq=\left(\frac{\sigma \pi R^2}{2}\right)\left(2.\frac{4R}{3\pi }\right)\\ \left(\because \text{distance of com from surface either upwards or downwards}, a=\frac{4R}{3\pi }\right)\end{array}$

[Step 2]:  $P=\frac{4R^3\sigma }{3}$
[Step 3]:  $f=Ma$
[Step 4]:

  $\begin{array}{l}{\tau }_E-{\tau }_f=I\alpha \\ \text{}\Rightarrow P{E}_0-fR=I\alpha \\ \text{}\Rightarrow \frac{4R^3\sigma }{3}{E}_0-MaR=\frac{M{R}^2\alpha }{2}\text{}\\ \Rightarrow \frac{4R^3\sigma }{3}{E}_0=MaR+\frac{M{R}^2\left(\frac{a}{R}\right)}{2}\left(\because For pure rolling, \alpha =\frac{a}{R}\right)\text{}\\ \Rightarrow \frac{4R^3\sigma }{3}{E}_0=\frac{3MRa}{2}\text{}\end{array}$
[Step 5]:  $For pure rolling, \alpha =\frac{a}{R}$
[Step 6]: For pure rolling  $a=\frac{8}{9}\frac{\sigma E_0{R}^2}{M}$