Mary Jackson

Answered

2022-01-15

A thin, horizontal, 11-cm-diameter copper plate is charged to -2.9 nC. Assume that the electrons are uniformly distributed on the surface

What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:$1.7\cdot {10}^{4}\frac{N}{C}$

What is the direction of the electric field 0.1 mm above the center of the top surface of the plate?

Answer: toward the plate

What is the strength of the electric field at the plates

What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

What is the direction of the electric field 0.1 mm above the center of the top surface of the plate?

Answer: toward the plate

What is the strength of the electric field at the plates

Answer & Explanation

Jim Hunt

Expert

2022-01-16Added 45 answers

Step 1 strength of electric field

Strength of electric field is calculated using charge density. It is defined as the charge contained over a surface area.

$\text{Charge density}=\frac{\text{Charge}}{\text{surface area}}$

$\text{Electric field intensity}=\frac{\text{charge density}}{2{\xi}_{0}}$

Where,$\xi}_{0}=8.85\times {10}^{-12}\frac{F}{m$

Step 2 calculate electric field strength

1.strength of electric field above the center of top surface is

Charge density is,

$\sigma =\frac{-2.9\times {10}^{-9}}{\pi \times {\left(0.055\right)}^{2}}$

$\sigma =-3.08\times {10}^{-7}\frac{C}{{m}^{2}}$

for distance 0.1 mm is very small compared to plates diameter. So assuming plate as infinite sheet of charge.

Strength of electric field,

$E=\frac{\sigma}{2{\xi}_{0}}$

$E=\frac{-3.08\times {10}^{-7}}{2\times 8.85\times {10}^{-12}}$

$E=-1.74\times {10}^{6}\frac{N}{C}$

2. negative sign indicates that direction of field is towards the sheet.

3. Net charge on the center of mass of plate is zero, so electric field is zero and so, is the strength of electric field.

$E=0\frac{N}{C}$

4. Strength of electric field 0.1 mm below is just the same as above the top surface of plate.

As copper is a conductor, it distributes charge itself across it,$E=1.74\times {10}^{4}\frac{N}{C}$

5. Direction of field remains same below the plate

Strength of electric field is calculated using charge density. It is defined as the charge contained over a surface area.

Where,

Step 2 calculate electric field strength

1.strength of electric field above the center of top surface is

Charge density is,

for distance 0.1 mm is very small compared to plates diameter. So assuming plate as infinite sheet of charge.

Strength of electric field,

2. negative sign indicates that direction of field is towards the sheet.

3. Net charge on the center of mass of plate is zero, so electric field is zero and so, is the strength of electric field.

4. Strength of electric field 0.1 mm below is just the same as above the top surface of plate.

As copper is a conductor, it distributes charge itself across it,

5. Direction of field remains same below the plate

Juan Spiller

Expert

2022-01-17Added 38 answers

Step 1

Given data:

Diameter of the copper plate,$d=14cm=0.14m$

Charge,$Q=-2.8nC=-2.8\times 10-9C$

Distance,$x=0.1mm=0.1\times 10-3m$

Step 2

To determine:

C. Electric field strength at the plates

Given data:

Diameter of the copper plate,

Charge,

Distance,

Step 2

To determine:

C. Electric field strength at the plates

alenahelenash

Expert

2022-01-23Added 366 answers

Complete Question A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate? Answer:
The values is $E=248.2N/C$
Explanation: From the question we are told that.
The diameter is $d=12cm=0.12m$
The distance from the center is $k=0.1mm=1\cdot {10}^{-4}m$
Generally the radius is mathematically represented as $r=\frac{d}{2}$
$\Rightarrow r=\frac{0.12}{2}$
$\Rightarrow r=0.06m$
Generally electric field is mathematically represented as
$E=\frac{Q}{2{\xi}_{0}}[1-\frac{k}{\sqrt{{r}^{2}+{k}^{2}}}]$
substituting values
$E=\frac{4.4\cdot {10}^{-9}}{2\cdot (8.85\cdot {10}^{-12})}[1-\frac{(1.00\cdot {10}^{-4})}{\sqrt{(0.06{)}^{2}+(1.0\cdot {10}^{-4}{)}^{2}}}]$
$E=248.2N/C$

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