Relation between Electric Potential and Electric Field

[Alan I]

Homework Statement

In a certain region of space, the electric field along the x-axis is given by:

E = 1.2x - 3.06, where E is in N/C and x is in meters.

If you set the electric potential equal to zero at (4.39 m,0), find the electric potential, in V, at the point (7.22 m,0).

Homework Equations

V = - ∫E * dl

The Attempt at a Solution

[/B]
V = - ∫ (1.2X-3.06) dx

⇒ V = - [1.2X²/2 - 3.06X]₀^2.83

⇒V=3.85 → which is http://www.dabur.com/odomos/images/wrong_sign.jpg ...any suggestions?

 

[Orodruin]

You have the wrong integration boundaries and you are missing the term due to the potential at x = 4.39 m.

 

[rude man]

Homework Statement

In a certain region of space, the electric field along the x-axis is given by:
E = 1.2x - 3.06, where E is in N/C and x is in meters.
If you set the electric potential equal to zero at (4.39 m,0), find the electric potential, in V, at the point (7.22 m,0).

First, find the potential difference between the two points.
Then set the one to zero & get the other.

 

[Alan I]

First, find the potential difference between the two points.
Then set the one to zero & get the other.

potential difference between the two points: V_ba = - ∫_4.39^7.22 (1.2X-3.06) dx

⇒ V_ba = -[1.2X²/2-3.06X]_4.39^7.22
= -(9.184+1.870)
= -11.054
⇒-11.054 = V_b - V_a

V_a=0

⇒V_b = -11.054 V

does that make more sense?

 

[rude man]

potential difference between the two points: V_ba = - ∫_4.39^7.22 (1.2X-3.06) dx

⇒ V_ba = -[1.2X²/2-3.06X]_4.39^7.22
= -(9.184+1.870)
= -11.054
⇒-11.054 = V_b - V_a

V_a=0

⇒V_b = -11.054 V

does that make more sense?

A whole lot more!

P.S. I didn't check the math.