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2 years ago

a charged partocle produces an electric field with a magnitude of 2.0 N/C at a point that is 50cm away from the particle

Physics

1 answer:

zheka24 [161]2 years ago

4 0

The charge on the particle is 5.6 × 10⁻¹¹ C.

<h3>Calculation:</h3>

The magnitude of an electric field produced by a charge is given by:

E = q/ 4πε₀r²

where,

E = electric field

q = charge

r = distance

1/4πε₀ = 8.99 × 10⁹ Nm²/C²

Given,

E = 2.0 N/C

r = 50 cm = 0.5 m

To find,

q =?

Put the values in the above equation:

E = q/ 4πε₀r²

q = E (4πε₀r²)

q = 2.0 × (0.50²)/ 8.99 × 10⁹

q = 5.6 × 10⁻¹¹ C

Therefore, the particle has a charge of 5.6 × 10⁻¹¹ C.

<h3>What is an electric field?</h3>

The physical field that surrounds each electric charge and acts to either attract or repel all other charges in the field is known as an electric field. Electric charges or magnetic fields with different amplitudes are the sources of electric fields.

I understand the question you are looking for is this:

A charged particle produces an electric field with a magnitude of 2.0 N/C at a point that is 50 cm away from the particle. What is the magnitude of the particle's charge?

Learn more about electric field here:

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Two thin parallel conducting plates are placed 2.0 cm apart. Each plate is 2.0 cm on a side; one plate carries a net charge of 8

azamat

Gauss's law and charges of the same sign repel allows us to find the results for the questions about the charged plates are:

The charge inside the plates is zero.

The field in the middle of the plates is: E = 2.26 10⁶ N/C

<h3>Gauss's law.</h3>

Gauss's law says that the electric flux through a Gaussian surface is proportional to the charge inside it.

Ф = ∫ E . dA = $\frac{q_{int}}{\epsilon_o }$

where Ф is the flux, E the electric field, A the area, $q_{int}$ the charge inside the surface.

They indicate that we have two metallic plates with a charge of 80 μC = 80 10⁻⁶ C in each one, since the plate is metallic, the electrons are free to move in it and repel each other, therefore the ones that are farthest from each other are placed, this is concentrated on the surface of the metal plate, therefore the charge inside the surface is zero.

Let's use Gauss's law to find the electric field, we define a Gaussian surface with a cylinder base parallel to the plate, in this case the field created by the charge is parallel to the normal of the surface of the plates.

2 E A = $\frac{q_{int}}{\epsilon_o}$

The two comes from the fact that the electric field is emitted towards both sides of the plate.

The charge density on each plate is:

σ = q A

Let's substitute.

E A = $\frac{\sigma A}{2 \ \epsilon_o}$

The electric field is a vector magnitude, so vector addition must be used, see attached for the direction of the electric field.

$R_{total} = E_1+E_2$

$E_{total} = \frac{\sigma}{\epsilon_o}$

Let's calculate.

The charge density.

$\sigma = \frac{q}{l^2}$

$\sigma = \frac{ 80 \ 10^{-6} } { 2.0 \ 2.0}$

σ = 20 10⁻⁶ C

The total electric field.

E = $\frac{20 \ 10^{-6} }{8.85 \ 10^{-12} }$

E = 2.26 10⁶ N/C

In conclusion, using Gauss's law and that charges of the same sign repel each other, we can find the result for the questions about the charged plates: