Effect of radius changes on electric fields and potential difference?

In summary, the conversation discusses the influence of radius on electric potential and electric fields. It is noted that electric potential is affected by the radius doubling due to its proportionality to 1/r, while electric fields remain unchanged due to the cancellation of equal charges on both sides. The conversation concludes with a clarification on the difference between a scalar and vector quantity.
  • #1
mirandab17
40
0
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Hello!

Okay so I understand that electric potential:

V = kQ/r

...must be influenced by the radius doubling because it would make the potential energy half of what it originally was because of the proportionality law, v is proportional to 1/r.

With electric fields though, how can there possibly be no change? The formula is

E = kQ/r^2

...meaning if should be influenced as well?

The answer is c, btw.
 
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  • #2
So do the math - calculate the field before and after the change in position.
Note: E is a vector.
 
  • #3
Oh! Right!

Since E is a vector, and the distance and charges on both sides are equal, then they always simply cancel out to zero. Whereas with electric potential, a scalar quantity, it is not affected by direction, merely magnitude, in which case both are positive, so the radius change will definitely affect it.

Thanks bud!
 
  • #4
No worries - that "Oh! Right!" feeling is what I was going for :)
 
  • #5


Hello! You are correct in understanding that changes in radius will have an effect on both electric potential and electric fields. As the radius increases, the electric potential will decrease according to the inverse proportionality relationship with the radius. This means that the potential energy will decrease as well. For electric fields, the strength of the field will also decrease with an increase in radius, as shown in the formula E = kQ/r^2. This is because the electric field is also inversely proportional to the square of the radius. So, doubling the radius will result in the electric field being reduced by a factor of 4. It is important to note that while the potential and field strength will change, the potential difference between two points will remain the same as long as the distance between those points remains constant. This is why the correct answer is c, as the potential difference is not affected by changes in radius. I hope this helps clarify any confusion!
 

Related to Effect of radius changes on electric fields and potential difference?

1. How does changing the radius affect the strength of an electric field?

Changing the radius of a charged object does not directly affect the strength of the electric field. The strength of an electric field is determined by the magnitude of the charge and the distance from the charge.

2. Does increasing the radius of a charged object increase the potential difference?

No, increasing the radius of a charged object does not directly affect the potential difference. The potential difference is determined by the work done to move a unit charge from one point to another, which is not affected by the radius of the charge.

3. How does changing the radius of a charged object affect the electric potential energy?

Changing the radius of a charged object does not directly affect the electric potential energy. The potential energy is determined by the product of the charge and the potential difference, which is not affected by the radius of the charge.

4. Can changing the radius of a charged object change the direction of the electric field?

Changing the radius of a charged object does not change the direction of the electric field. The electric field always points away from positive charges and towards negative charges, regardless of the radius of the charge.

5. How does the electric field change when the radius of a charged object is doubled?

The electric field strength decreases by a factor of four when the radius of a charged object is doubled. This is because the electric field strength is inversely proportional to the square of the distance from the charge, so doubling the distance decreases the field strength by a factor of four.

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