Question

A parallel-plate capacitor has plate area 25⋅0 cm2 and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.

Answer 1

Answer:

I don't know the answer

Explanation:

I think 202v and 2+

Answer 2

The extra charge given by the battery to the positive plate is equal to the $1.33\times 10-10 C$

Explanation:

The capacitance of capacitor is given by C = €A/d

Where € = dielectric constant of the capacitor

A= Area of the plates

d= distance between the plates

Given,  

Area of the plates is =  $25 \cdot 0 \mathrm{cm} 2=25 \times 10-4$

Distance between the plates = d = 2 mm = $2\times 10-3$

Potential difference between the plates = V = 12 V

Substitute the above values in to the formula

$C=8.85 \times 10-12 \times 25 \times 10-4 / 2 \times 10-3 =11.06 \times 10-12 \mathrm{F} We know that Charge Q=C X V Q=11.06 \times 10-12 \times 12 =1.33 \times 10-10 c$

Considering the second statement the separation between the plates is decreased and capacitance we can consider as C1

$\Rightarrow \quad C 1=8.85 \times 10-12 \times 25 \times 10-4 / 1 \times 10-3 \Rightarrow 22.12 \times 10-12 F$

We know that Charge Q = $C \times V$

$\begin{aligned}&Q 1=22.12 \times 10-12 \times 12\\&=2.65 \times 10-10 c\end{aligned}$

For calculating the extra charge Q2 = Q1 – Q

$\begin{aligned}&\Rightarrow(2.65 \times 10-10-1.33 \times 10-10)\\&\Rightarrow \quad 1.33 \times 10-10\end{aligned}$

Therefore, the extra charge given by the battery to the positive plate is equal to the $1.33 \times 10-10 C.$