What Determines Electric Field Strengths of Two Charged Rods

In summary, the problem at hand involves two 10cm long thin rods, one made of glass and one made of plastic, that are uniformly charged to 10nC and -10nC respectively. They are placed 4cm apart and the question asks for the electric field strength at distances of 1cm, 2cm, and 3cm from the glass rod along the line connecting the midpoints of the two rods. The formula E= η/ε0 = (4πK⋅Q) / A is not applicable in this scenario as it is not describing a capacitor. Instead, the electric field due to a finite line of charge must be calculated and then superimposed for both rods. Additional
  • #1
miyayeah
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Homework Statement


The problem I encountered goes like this: A 10cm long thin glass rod uniformly charged to 10nC and a 10cm long thin plastic rod uniformly charged to -10nC are placed side by side 4cm apart. What are the electric field strengths E1 to E3 at distances 1cm, 2cm, and 3cm from the glass rod along the line connecting the midpoints of the two rods?

Homework Equations


E= η/ε0 = (4πK⋅Q) / A

The Attempt at a Solution


I am not quite confused about getting the answer to this problem but as to how I should approach the problem. I thought that the question was describing a capacitor, since the two rods are charged equally but oppositely. But when I tried to use the formula (as shown above), I realized I have no way to find the area. So, I realized that it was not a capacitor. How would I know, aside from looking at the formula, that the question is not describing a capacitor? Or when it describes a capacitor?
 
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  • #2
To have a capacitor, you need two conductors near one another carrying equal and opposite charges. That's not what you have here because neither glass nor plastic can be considered as conductors. You need to find the electric field due to finite line of charge and superimpose the two fields.
 
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  • #3
kuruman said:
To have a capacitor, you need two conductors near one another carrying equal and opposite charges. That's not what you have here because neither glass nor plastic can be considered as conductors. You need to find the electric field due to finite line of charge and superimpose the two fields.
Thank you for your explanation. Which formula should I use if I am solving for the electric field due to finite line of charge?
 

1. How do you calculate the capacitance value of a capacitor?

The capacitance value of a capacitor can be calculated by dividing the charge stored on one of its plates by the potential difference (voltage) between the plates. This is represented by the equation C = Q/V, where C is the capacitance in farads (F), Q is the charge in coulombs (C), and V is the voltage in volts (V).

2. What is the unit of measurement for capacitance?

The unit of measurement for capacitance is farads (F), named after the English physicist Michael Faraday. In practical applications, capacitance is often measured in microfarads (μF) or picofarads (pF), which are smaller units of measurement.

3. How do you determine the maximum voltage a capacitor can withstand?

The maximum voltage a capacitor can withstand, also known as its breakdown voltage, is determined by the material and design of the capacitor. This information is typically provided by the manufacturer and can be found in the capacitor's datasheet. It is important to never exceed the maximum voltage rating of a capacitor to avoid damaging it.

4. What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by several factors, including the distance between the plates, the surface area of the plates, and the type of dielectric material between the plates. The capacitance increases with a decrease in distance between the plates, an increase in surface area of the plates, and a higher dielectric constant of the material. Additionally, the capacitance is also affected by temperature and frequency.

5. How do you measure the capacitance of a capacitor?

The capacitance of a capacitor can be measured using a capacitance meter or by using a multimeter set to the capacitance function. When using a multimeter, it is important to ensure that the capacitor is fully discharged before measuring and to select the appropriate range on the multimeter. The measured capacitance may differ slightly from the rated value due to factors such as tolerance and temperature.

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