Question

Two charges +3.2 x 10-19 C and -3.2 x 10-19C kept 2.4 Å apart form a dipole. If it is kept in uniform electric field of intensity 4 x 105 V/m then. what will be its electrical energy in stable equilibrium.

Answer 1

[AI-generated answer — Explain Bot]

To find the electrical energy in stable equilibrium of the dipole in a uniform electric field, we can use the equation:

Electric Potential Energy (U) = -pEcosθ

Where:
- U is the electric potential energy
- p is the dipole moment
- E is the electric field intensity
- θ is the angle between the dipole moment and the electric field

In this case, the charges are +3.2 x 10^-19 C and -3.2 x 10^-19 C, which are equal in magnitude. Thus, the dipole moment (p) can be calculated as the product of the charge (q) and the distance between them (d):

p = q * d

Where:
- q is the charge
- d is the distance between the charges

Given that the charges are 2.4 Å (angstrom) apart, we need to convert it to meters:

1 Å = 1 x 10^-10 m

So, 2.4 Å = 2.4 x 10^-10 m

Now we can calculate the dipole moment:

p = (3.2 x 10^-19 C) * (2.4 x 10^-10 m)

Next, we need to calculate the angle θ between the dipole moment and the electric field. For a dipole aligned with the field, θ = 0 degrees (cos 0 = 1). Hence, the formula simplifies to:

Electric Potential Energy (U) = -pE

Substituting the values:

U = -(3.2 x 10^-19 C * 2.4 x 10^-10 m) * (4 x 10^5 V/m)

Now, we can calculate the electric potential energy in stable equilibrium.