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.

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Khoso123 Khoso123 · Answer 1 · 2020-02-06T05:41:50+01:00

Answer:

(a) [tex]E=200000Volts/meter[/tex]

(b) [tex]E=200000Volts/meter[/tex]

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

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

As we know that

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

So Electric potential is given as:

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

For (b) Potential at distance=2.00mm

First we need to find the capacitance

So

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

As we know that

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

So Electric potential is given as:

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