Math / Algebra

Question3. Consider two parallel plates $2.00$ mm apart with a potential difference of $240$ V and a positively charged upper plate. A charged oil droplet with a mass of $5.88 \times 10^{-10}$ kg is suspended between the plates. Determine the sign and magnitude of the electric charge on the oil droplet, and calculate the electron deficiency or excess. $0.002$ m

Studdy Solution

STEP 1

What is this asking? We've got a tiny oil droplet just chilling between two electrically charged plates, and we need to figure out what kind of charge it has, how strong that charge is, and whether it's missing or has extra electrons! Watch out! Don't mix up the forces!
Gravity pulls down, and the electric field can push or pull depending on the charge.
Also, remember that electron deficiency means a positive charge, and excess means a negative charge.

STEP 2

1. Analyze the forces
2. Calculate the electric field
3. Determine charge
4. Calculate electron difference

STEP 3

Alright, so our little oil droplet is just hanging there, not moving up or down.
That means the forces on it are balanced.
The downward force of gravity must be perfectly matched by an upward electric force.

STEP 4

Since the upper plate is positively charged and the droplet is being pushed up, away from it, the droplet must be positively charged too!
Like charges repel!

STEP 5

We know that the force of gravity is given by $F_g = m \cdot g$ , where $m$ is the mass and $g$ is the acceleration due to gravity ( $9.81 \, m/s^2$ ).
The electric force is given by $F_e = q \cdot E$ , where $q$ is the charge we're looking for, and $E$ is the electric field strength.
Since these forces are balanced, we can set them equal: $m \cdot g = q \cdot E$ .

STEP 6

The electric field strength between two parallel plates is given by $E = \frac{V}{d}$ , where $V$ is the potential difference and $d$ is the distance between the plates.

STEP 7

Let's plug in the values: $E = \frac{240 \, V}{0.002 \, m} = 120000 \, V/m$ .
So, the electric field strength is a whopping 120,000 V/m!

STEP 8

Now we can go back to our balanced forces equation: $m \cdot g = q \cdot E$ .
We want to solve for $q$ , the charge, so we rearrange it to $q = \frac{m \cdot g}{E}$ .

STEP 9

Plugging in our values: $q = \frac{5.88 \times 10^{-10} \, kg \cdot 9.81 \, m/s^2}{120000 \, V/m} \approx 4.80 \times 10^{-14} \, C$ .
So, the charge on the oil droplet is approximately $4.80 \times 10^{-14} \, C$ , and it's positive, just like we figured out earlier!

STEP 10

The elementary charge, which is the charge of a single proton or electron (ignoring the sign), is $e = 1.60 \times 10^{-19} \, C$ .
To find the electron deficiency, we divide the total charge by the elementary charge: $\text{Number of missing electrons} = \frac{q}{e}$ .

STEP 11

Let's calculate: $\text{Number of missing electrons} = \frac{4.80 \times 10^{-14} \, C}{1.60 \times 10^{-19} \, C} = 300000$ .
That means the oil droplet is missing a whopping 300,000 electrons!

STEP 12

The oil droplet has a positive charge of approximately $4.80 \times 10^{-14} \, C$ , and it has a deficiency of 300,000 electrons.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the forces
2. Calculate the electric field
3. Determine charge
4. Calculate electron difference

STEP 3

Alright, so our little oil droplet is just hanging there, not moving up or down.
That means the forces on it are balanced.
The downward force of gravity must be perfectly matched by an upward electric force.

STEP 4

Since the upper plate is positively charged and the droplet is being pushed up, away from it, the droplet must be positively charged too!
Like charges repel!

STEP 5

STEP 6

The electric field strength between two parallel plates is given by $E = \frac{V}{d}$ , where $V$ is the potential difference and $d$ is the distance between the plates.

STEP 7

Let's plug in the values: $E = \frac{240 \, V}{0.002 \, m} = 120000 \, V/m$ .
So, the electric field strength is a whopping 120,000 V/m!

STEP 8

Now we can go back to our balanced forces equation: $m \cdot g = q \cdot E$ .
We want to solve for $q$ , the charge, so we rearrange it to $q = \frac{m \cdot g}{E}$ .

STEP 9

Plugging in our values: $q = \frac{5.88 \times 10^{-10} \, kg \cdot 9.81 \, m/s^2}{120000 \, V/m} \approx 4.80 \times 10^{-14} \, C$ .
So, the charge on the oil droplet is approximately $4.80 \times 10^{-14} \, C$ , and it's positive, just like we figured out earlier!

STEP 10

The elementary charge, which is the charge of a single proton or electron (ignoring the sign), is $e = 1.60 \times 10^{-19} \, C$ .
To find the electron deficiency, we divide the total charge by the elementary charge: $\text{Number of missing electrons} = \frac{q}{e}$ .

STEP 11

Let's calculate: $\text{Number of missing electrons} = \frac{4.80 \times 10^{-14} \, C}{1.60 \times 10^{-19} \, C} = 300000$ .
That means the oil droplet is missing a whopping 300,000 electrons!

STEP 12

The oil droplet has a positive charge of approximately $4.80 \times 10^{-14} \, C$ , and it has a deficiency of 300,000 electrons.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the forces2. Calculate the electric field3. Determine charge4. Calculate electron difference

STEP 3

Alright, so our little oil droplet is just hanging there, not moving up or down.That means the forces on it are **balanced**.The downward force of **gravity** must be perfectly matched by an upward **electric force**.

STEP 4

Since the upper plate is positively charged and the droplet is being pushed *up*, away from it, the droplet *must* be positively charged too!Like charges repel!

STEP 5

STEP 6

The electric field strength between two parallel plates is given by E=VdE = \frac{V}{d}E=dV​, where VVV is the **potential difference** and ddd is the **distance** between the plates.

STEP 7

Let's plug in the values: E=240 V0.002 m=120000 V/mE = \frac{240 \, V}{0.002 \, m} = 120000 \, V/mE=0.002m240V​=120000V/m.So, the electric field strength is a whopping **120,000 V/m**!

STEP 8

Now we can go back to our balanced forces equation: m⋅g=q⋅Em \cdot g = q \cdot Em⋅g=q⋅E.We want to solve for qqq, the charge, so we rearrange it to q=m⋅gEq = \frac{m \cdot g}{E}q=Em⋅g​.

STEP 9

STEP 10

STEP 11

STEP 12

The oil droplet has a positive charge of approximately 4.80×10−14 C4.80 \times 10^{-14} \, C4.80×10−14C, and it has a deficiency of 300,000 electrons.

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1. Analyze the forces
2. Calculate the electric field
3. Determine charge
4. Calculate electron difference

Alright, so our little oil droplet is just hanging there, not moving up or down.
That means the forces on it are balanced.
The downward force of gravity must be perfectly matched by an upward electric force.

Since the upper plate is positively charged and the droplet is being pushed up, away from it, the droplet must be positively charged too!
Like charges repel!

The electric field strength between two parallel plates is given by $E = \frac{V}{d}$ , where $V$ is the potential difference and $d$ is the distance between the plates.

Let's plug in the values: $E = \frac{240 \, V}{0.002 \, m} = 120000 \, V/m$ .
So, the electric field strength is a whopping 120,000 V/m!

Now we can go back to our balanced forces equation: $m \cdot g = q \cdot E$ .
We want to solve for $q$ , the charge, so we rearrange it to $q = \frac{m \cdot g}{E}$ .

The oil droplet has a positive charge of approximately $4.80 \times 10^{-14} \, C$ , and it has a deficiency of 300,000 electrons.