In the following formula, you can figure out relationships between variables, even though you may not know (yet) the meaning of these variables.
The electric potential V created at a distance of r from a point source is given by 
V = 
C
r
,
 and the electric field E is given by 
E = 
C
r2
,
 where C is a constant that depends on the source. What would be the relative change in V and in E if the distance r is doubled? 

Let's analyze how the electric potential \( V \) and the electric field \( E \) change when the distance \( r \) is doubled.

### Given Formulas
1. The electric potential \( V \) is given by:
   \[
   V = \frac{C}{r}
   \]
2. The electric field \( E \) is given by:
   \[
   E = \frac{C}{r^2}
   \]
where \( C \) is a constant.

### Change in Electric Potential \( V \)
If the distance \( r \) is doubled, the new distance becomes \( 2r \). The new potential \( V' \) is:
\[
V' = \frac{C}{2r}
\]
Now, compare \( V' \) with the original potential \( V \):
\[
\frac{V'}{V} = \frac{\frac{C}{2r}}{\frac{C}{r}} = \frac{1}{2}
\]
This means the electric potential \( V \) is reduced by half when the distance \( r \) is doubled.

### Change in Electric Field \( E \)
If the distance \( r \) is doubled, the new electric field \( E' \) is:
\[
E' = \frac{C}{(2r)^2} = \frac{C}{4r^2}
\]
Now, compare \( E' \) with the original field \( E \):
\[
\frac{E'}{E} = \frac{\frac{C}{4r^2}}{\frac{C}{r^2}} = \frac{1}{4}
\]
This means the electric field \( E \) is reduced to a quarter of its original value when the distance \( r \) is doubled.

### Summary of Changes
- **Electric Potential \( V \)**: When \( r \) is doubled, \( V \) becomes \( \frac{1}{2} \) of its original value.
- **Electric Field \( E \)**: When \( r \) is doubled, \( E \) becomes \( \frac{1}{4} \) of its original value.

Would you like further details or explanations on any part of this? Here are some related questions you might find interesting:

1. What happens to \( V \) and \( E \) if the distance \( r \) is halved?
2. How does the constant \( C \) affect the electric potential and field?
3. What are some real-life applications of these relationships?
4. How do the units of \( V \) and \( E \) relate to the physical quantities they represent?
5. What is the significance of the inverse square law in physics?

**Tip:** Understanding how physical quantities scale with distance is crucial in fields like electromagnetism and gravity. It often reveals the nature of forces and potentials in various physical systems.

Explore how the electric potential V and electric field E change when the distance r from a point source is doubled. Understand the impact of doubling distance on electric potential and electric field with detailed explanations and formulas.

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