Question

Two 2.00 cm * 2.00 cm plates that form a parallel-plate capacitor are charged to { 0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is (a) 1.00 mm and (b) 2.00 mm.

Answer 1

Answer:

(a) $E=200000Volts/meter$

(b) $E=200000Volts/meter$

Explanation:

Given data

Area A=4.0 cm²=0.0004 m²

Electric permittivity ε=8.854×10⁻¹² farads/meter

Charge q=0.708 nC

To find

(a) Potential at distance=1.00mm

(b) Potential at distance=2.00mm

Solution

For (a) Potential at distance=1.00mm

First we need to find the capacitance

So

$C=E*A/D\\C=8.854*10^{-12}*(0.0004/1.00*10^{-3} )\\ C=3.5416*10^{-12}farads$

As we know that

$Q=CV\\V=Q/C\\V=(0.708*10^{-9}C )/3.5416*10^{-12}farads\\V=200Volts$

So Electric potential is given as:

$E=V/d\\E=(200V)/(1*10^{-3}m )\\E=200000Volts/meter$

For (b) Potential at distance=2.00mm

First we need to find the capacitance

So

$C=E*A/D\\C=8.854*10^{-12}*(0.0004/2.00*10^{-3} )\\ C=1.7708*10^{-12}farads$

As we know that

$Q=CV\\V=Q/C\\V=(0.708*10^{-9}C )/1.7708*10^{-12}farads\\V=400Volts$

So Electric potential is given as:

$E=V/d\\E=(400V)/(2.0*10^{-3}m )\\E=200000Volts/meter$