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Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Respuesta :

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

[tex]Electric FIeld = \frac{Change\: in\: Potential}{Distance}[/tex]

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance=[tex]0.8 \times 10^{-2} m[/tex]

So, Electric field is

[tex]Electric\: field=\frac{3.8V}{0.8 \times 10^{-2}m }[/tex]

[tex]Electric\: field=475 V/m[/tex]

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

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