Question

A capacitor has a capacitance of $27.0microfarads$.
If we triple the area of the plates of the capacitance and cut the distance between the plates to $1/3$ of its original value, what is the new capacitance of the capacitor?

• A. $243microfarads$
• B. $9.0microfarads$
• C. $3.0microfarads$
• D. $81.0microfarads$
• E. $27.0microfarads$

Answer 1

Answer: (A)

$243microfarads$

The capacitance of a parallel plate capacitor is given by ,

$C=\epsilon A/d$ ,

where $A=$ area of plate ,

$d=$ distance between plates ,

given $C=27.0\mu F$ ,

therefore $27.0=\epsilon A/d$ ............................eq1

now when $area=3A$ and $distance=d/3$ ,

then $C^{\prime }=\epsilon \times 3A/\left(d/3\right)=9\epsilon A/d$ ..........................eq2

dividing eq2 by eq1 ,

$C^{\prime }/27.0=9$ ,

or $C^{\prime }=27.0\times 9=243\mu F$