Electric Field of a Conducting Sphere

In summary, the conversation discusses how to calculate the electric field at a certain distance from the center of a charged conducting sphere. The formula E(r)=1/4∏ε_0 * qr/R^3 is mentioned, and two different methods are attempted to solve for the electric field at a distance of 0.612cm from the center of the sphere. There is also a question about whether the formula applies to the field outside the sphere and a discussion about the relationship between electric field and distance. The final answer of 1440.874 N/C is given, but it is noted that this answer was considered incorrect on the online homework site.
  • #1
Yosty22
185
4

Homework Statement



At a distance of 0.206cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485N/C . What is the electric field 0.612cm from the center of the sphere?

Homework Equations



E(r)=1/4∏ε_0 * qr/R^3
where r is radius of the Gaussian surface and R is the radius of the sphere

The Attempt at a Solution



I tried this two different ways, and they were both the same answer (and both wrong). Firstly, I used the given E-field (485N/C) and plugged it into the above equation and solved for q. Once I solved for q, I used that q with a new distance from the center of the sphere to solve for the new E Field at that position. When I did that, I got that the electric field should equal 1440.874 N/C. I plugged this into my online homework site, and it said I was wrong.

Since that way seemed to be wrong, I tried to just set it up as a ratio. I said that:

(485N/C)/0.00206m = (x N/C)/.00612m

Solving for x, which represents the E Field, I again got 1440.874 N/C.

Any ideas where I might have gone wrong?
 
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  • #2
Yosty22 said:

Homework Statement



At a distance of 0.206cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485N/C . What is the electric field 0.612cm from the center of the sphere?

Homework Equations



E(r)=1/4∏ε_0 * qr/R^3
where r is radius of the Gaussian surface and R is the radius of the sphere

Does this formula apply to the field outside the sphere?

I tried this two different ways, and they were both the same answer (and both wrong). Firstly, I used the given E-field (485N/C) and plugged it into the above equation and solved for q.

How? Using the formula above?

(485N/C)/0.00206m = (x N/C)/.00612m

Is electric field inversely proportional to distance?
 

Related to Electric Field of a Conducting Sphere

1. What is the Electric Field of a Conducting Sphere?

The Electric Field of a Conducting Sphere is a measure of the strength and direction of the electric force at any point on the surface of a spherical conductor. It is defined as the force per unit charge at that point.

2. How is the Electric Field of a Conducting Sphere calculated?

The Electric Field of a Conducting Sphere can be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere to the point where the electric field is being measured.

3. What is the direction of the Electric Field of a Conducting Sphere?

The Electric Field of a Conducting Sphere is radial, meaning it points directly away from or towards the center of the sphere. The direction depends on the charge of the sphere, positive or negative, and the location of the point where the field is being measured.

4. How does the charge of the sphere affect the Electric Field?

The charge of the sphere directly affects the strength of the Electric Field. As the charge increases, the Electric Field also increases. Additionally, the direction of the Electric Field is determined by the charge of the sphere.

5. What are some real-world applications of the Electric Field of a Conducting Sphere?

The Electric Field of a Conducting Sphere has many practical applications, including in electrostatics experiments, electronic devices such as capacitors and antennas, and in lightning protection systems. It is also used in industries such as telecommunications, aerospace, and energy production.

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