A parallel-plate capacitor has plates of area 200 cm 2 and separation between the plates 1 mm- The potential difference developed if a charge of 1 nC i-e-1 - 10-9 C is given to the capacitor is something-then if the plate separation is now increased to 2 mm- what will be the new potential difference-

We know that Q/C=V nbsp; nbsp; nbsp; nbsp; nbsp; .......(i) And also C=ε_0 A/d Putting this in eq(i) V= 1 × 10^ 9×1×10^ 3/200×10^ 4×8.85×10^ 12=5. ...

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Standard XII

Physics

Potential and Potential Difference

A parallel-pl...

Question

A parallel-plate capacitor has plates of area $200 c m^{2}$ and separation between the plates $1 m m$ . The potential difference developed if a charge of $1 n C (i . e . 1 \times 10^{- 9} C)$ is given to the capacitor is something,then if the plate separation is now increased to $2 m m$ , what will be the new potential difference?

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Solution

We know that
$\frac{Q}{C} = V$ .......(i)
And also $C = \frac{ε_{0} A}{d}$
Putting this in eq(i)
$V = \frac{1 \times 10^{- 9} \times 1 \times 10^{- 3}}{200 \times 10^{- 4} \times 8.85 \times 10^{- 12}} = 5.65 V$
If plate separation $(d)$ is made $2 mm$ then potential difference $(V^{'})$ is given by
$V^{'} = \frac{Q}{C} = \frac{1 \times 10^{- 9} \times 2 \times 10^{- 3}}{200 \times 10^{- 4} \times 8.85 \times 10^{- 12}} = 11.3 V$

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We know that QC=V .......(i) And also C=ε0Ad Putting this in eq(i) V=1×10−9×1×10−3200×10−4×8.85×10−12=5.65 V If plate separation(d) is made 2 mm then potential difference (V′) is given by V′=QC=1×10−9×2×10−3200×10−4×8.85×10−12=11.3 V