Given:
The electric field intensity between the parallel plates is,
[tex]E_1[/tex]The distance between the plates is doubled and the charges are halved.
The area of the plates is tripled.
To find:
the new electric field
Explanation:
The electric field intensity between the parallel plates is,
[tex]E_1=\frac{Q}{A\epsilon}[/tex]Q is the charge on each plate and A is the area of each plate.
Now the field intensity is,
[tex]\begin{gathered} E_2=\frac{\frac{Q}{2}}{3A\times\epsilon} \\ =\frac{Q}{6A\epsilon} \\ =\frac{E_1}{6} \end{gathered}[/tex]Hence, the new electric field intensity is d.
[tex][/tex]