Question

A water molecule perpendicular to an electric field has 1.40×10−21 J more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is 6.2×10−30Cm.
What is the strength of the electric field?
Express your answer with the appropriate units.

Answer 1

Answer:

The strength of the electric field is 2.258 x 10⁸ N/C

Explanation:

Potential energy = -p*ECosθ

where;

d is the dipole moment

E is the electric field  

θ is the angle of inclination

→When the water molecules is perpendicular to the field, θ =90° and potential energy = p*ECosθ = d*ECos0 = 0

Total potential energy = 0 + 1.40×10⁻²¹J, since it is 1.40×10⁻²¹J more

→When the water molecules is aligned to the field, θ =0°

potential energy = -p*ECosθ

dipole moment, p = 6.2×10⁻³⁰Cm

                            = -(6.2×10⁻³⁰)*(E) Cos0

0 + 1.40×10⁻²¹ J = -(6.2×10⁻³⁰)*(E),

E = (1.40×10⁻²¹ J)/(6.2×10⁻³⁰)

E = 2.258 x 10⁸ N/C

Therefore, the strength of the electric field is 2.258 x 10⁸ N/C

Answer 2

The strength of the electric field is 2.3 × 10⁸ N/C.

Electric field strength

It is a quantitative expression of the strength of an electric field at a specific location.

Based on the given information,

• The dipole moment or P of the water molecule is 6.2 × 10⁻³⁰ cm.

• A water molecule perpendicular to an electric field has 1.40 × 10⁻²¹ J more potential energy in comparison to a water molecule aligned with the field.

Now the potential energy of a water molecule, when it is perpendicular,

PEp = PEcosФ

PEp = PEcos90°

PEp = 0

Now the potential energy when it is aligned to electric field is,

PEa = PE CosΦ

PEa = PE CosO°

PEa = PE

Now the change in potential energy is,

ΔV = PE (CosΦp - CosΦa)

ΔV = -PE

E = ΔV/P

E = -PE/P

E = 1.4 × 10⁻²¹ J/6.2 × 10⁻³⁰

E = 2.3 × 10⁸ N/C

Thus, the strength of the electric field is  2.3 × 10⁸ N/C.

Find out more information about electric field strength here:

https://brainly.com/question/12117399