A disc having uniform surface charge density +σ in upper half and −σ 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

A
$\frac{8 σ R^{2} E_{0}}{9 M}$
B
$\frac{9 σ R^{2} E_{0}}{8 M}$
C
$\frac{4 σ R^{2} E_{0}}{3 M}$
D
$\frac{3 σ R^{2} E_{0}}{4 M}$

Solution:

[Step 1]: Dipole moment

$\begin{array}{l} (\begin{array}{l} ∵ σ = \frac{q}{A} \\ \Rightarrow q = σ A_{h e m i s p h e r e} = \frac{σ π R^{2}}{2} \end{array}) \\ \Rightarrow P = 2 a q = (\frac{σ π R^{2}}{2}) (2. \frac{4 R}{3 π}) \\ (∵ distance of com from surface either upwards or downwards, a = \frac{4 R}{3 π}) \end{array}$

[Step 2]: $P = \frac{4 R^{3} σ}{3}$
[Step 3]: $f = M a$
[Step 4]:

$\begin{array}{l} τ_{E} - τ_{f} = I α \\ \Rightarrow P E_{0} - f R = I α \\ \Rightarrow \frac{4 R^{3} σ}{3} E_{0} - M a R = \frac{M R^{2} α}{2} \\ \Rightarrow \frac{4 R^{3} σ}{3} E_{0} = M a R + \frac{M R^{2} (\frac{a}{R})}{2} (∵ F o r p u r e r o l l i n g, α = \frac{a}{R}) \\ \Rightarrow \frac{4 R^{3} σ}{3} E_{0} = \frac{3 M R a}{2} \end{array}$
[Step 5]: $F o r p u r e r o l l i n g, α = \frac{a}{R}$
[Step 6]: For pure rolling $a = \frac{8}{9} \frac{σ E_{0} R^{2}}{M}$

Solution:

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