shelbysmith928
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Consider a uniform electric field E of magnitude E = 7.2 N/C. The line segment connecting points A and B in the field is perpendicular to the direction of the field, while the line segment connecting points C and B is parallel to the field. The line segments have equal lengths of d = 0.38 m.
Find the potential difference, in volts, between point A and point B.
Enter an expression for the potential difference between point C and point B in terms of E and d.
Calculate the potential difference, in volts, between point C and point B.

Respuesta :

lucidnightmares

a. Va-Vb= 0 volts

The two are perpendicular and therefore would have a difference of zero

b. Vc-Vb= -Ed

We already know V=Ed and in this case d is negative

c. -2.736 volts

(-7.2)*(.38)= -2.736

Plug in what was given for E and d into the equation for b

The problem is about the relation between electric field and potential difference. They both are related according to:

Vb -Va = - ∫ E× dl

The solution is:

a) V(B) - V(A)  = 0

c) V(C) - V(B) =  2.74 (V)

a) The segment AB is perpendicular to the lines of the electric field, then the point product is zero and V(B) - V(A) = 0

b) The segment CB is parallel to the lines of the electric field, and it is uniform, the:

V(C) - V(B) = - ∫₀⁰°³⁸ E × dl = E× d |₀⁰°³⁸ = - (7.2)×(0.38) = - 2.74 (V)

So the expression for the potential difference between point C and point B in terms of E and d is:

V(C) - V(B) = - ( E×d)

And finally     V(C) - V(B) =  2.74 (V)

Note: related problems  Brainly.com/question/23899758

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