Question

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface charge density 4.00 × 10^−12 C/m^2 . 1) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00 × 10^−2 m from the sheet?
Express your answer to three significant figures and include the appropriate units.

2) What is the speed of the electron when it is 3.00 × 10^−2 m from the sheet?
Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

Answer 2

Electric field is a physical field. The speed of the electron when it is 3.00 × 10^−2 m from the sheet is 1.86 x 10⁵ m/s.

What is an electric field?

An electric field is a physical field that applies forces on all the other charged particles in the field.

A.)

We know that the electric field due to the large sheet is given by the formula,

$\overrightarrow E = \dfrac{\sigma}{2\epsilon_o}$

Since the surface charge density is given as 4.00 × 10⁻¹² C/m², therefore, the electric field of the sheet can be written as,

$\overrightarrow E = \dfrac{4 \times 10^{-12}}{2\times 8.85 \times 10^{-12}}\\\\\overrightarrow E = 0.226\rm\ \ N/C$

The acceleration of an electron due to the electric field is given by,

$a=\dfrac{eE}{m}$

Substitute the value of the electric field E,

$a=\dfrac{1.6 \times 10^{-19} \times0.226}{9.1 \times 10^{-31}}\\\\a = 3.97 \times 10^{10}$

Now, the change in the position of the electron can be written as,

d = Initial Position - final position

d= 0.470 - (3 x 10⁻²)

d = 0.470 - 0.03

d = 0.44 m

Further, the work done can be written as,

$W = F\cdot d\\W = (ma)\cdot d\\W = 9.11 \times 10^{-31} \times 3.97 \times 10^{10}\times 0.44\\W = 1.58 \times 10^{-20}\rm\ J$

Thus, the work done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00 × 10⁻² m from the sheet is 1.58x10⁻²⁰ J.

B.)

We know that the work done can be written as the change in the kinetic energy of the electron. So the work done can be written as,

$W = \dfrac{1}{2}m(v_1^2- v_o^2)\\\\1.58 \times10^{-20} = \dfrac{1}{2}\times 9.1 \times 10^{-31}\times (v_1^2- 0^2)\\\\v_1=1.86 \times 10^5\rm\ m/s$

Hence, the speed of the electron when it is 3.00 × 10⁻² m from the sheet is 1.86 x 10⁵ m/s.

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