# Charge density - parallel plates

• pat666
In summary, two large parallel conducting plates with equal and opposite charges separated by 2.20 cm have a surface charge density of 47.0 nCm^-2. The magnitude of the electric field intensity between the plates is 530.8 N/C. The potential difference between the two plates is 116.8 V. The relationship between field and potential is V = Ed.
pat666

## Homework Statement

5. Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm.
a) If the surface charge density for each plate has a magnitude of 47.0 nCm^-2, what is the magnitude of the electric field intensity in the region between the plates?
b) What is the potential difference between the two plates? (2 marks)

## The Attempt at a Solution

The wording here confuses me a bit, is Electric field intensity something different to the electric field?
I've done E=4.7*10^-9/8.85E-12 =530.8N/C but is this the field intensity? Part b I am not sure about, I need some help for it to.

Thanks

Electric field and electric field intensity: same thing. (Check your arithmetic: is the charge density 4.7 or 47?)

For part b: You'll need to use the distance.

47, so E=5310.7N/C
Then V=kq/r but I don't know q so how would I do that?

pat666 said:
Then V=kq/r but I don't know q so how would I do that?
That formula is for a point charge, which is not relevant here. What's the general relationship between field and potential?

found it, V=Ed

so V=116.8V?

You got it.

Thanks

## What is charge density?

Charge density refers to the amount of electric charge per unit volume at a given point in space. It is typically measured in coulombs per meter cubed (C/m3).

## How is charge density calculated?

To calculate charge density, you divide the total electric charge by the volume it occupies. For example, if a parallel plate has a charge of +3 C and an area of 1 m2, the charge density would be 3 C/m3.

## What is the relationship between charge density and electric field?

Charge density and electric field are directly proportional. This means that as charge density increases, so does the electric field, and vice versa. This relationship is described by the equation E = ρ/ε0, where E is the electric field, ρ is the charge density, and ε0 is the permittivity of free space.

## How does charge density affect capacitance in parallel plates?

Charge density plays a critical role in determining the capacitance of a parallel plate capacitor. The higher the charge density, the higher the capacitance, as there is more charge available to store energy in the electric field between the plates. This relationship is described by the equation C = ε0A/d, where C is capacitance, ε0 is permittivity of free space, A is the area of the plates, and d is the distance between the plates.

## What is the effect of changing the distance between parallel plates on charge density?

As the distance between parallel plates decreases, the charge density increases. This is because the same amount of charge is now occupying a smaller volume, resulting in a higher charge density. Similarly, as the distance between plates increases, the charge density decreases.

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