Question

An electric dipole is in a uniform electric field of magnitude 8.50×104N/C. The charges in the dipole are separated by 1.10×10−10m, and the torque on the dipole when its dipole moment is perpendicular to the electric field is 6.60×10−26N⋅m. Calculate the magnitude of the charges that make up the dipole.

Answer 1

Answer:

Magnitude of charges = 0.705 x 10⁻²⁰ C

Explanation:

Magnitude of electric field = 850 x 10⁴ N/C

Distance between dipoles = r 1.10 x 10⁻¹⁰ m

Torque on dipole = 6.6 x 10⁻²⁶ N.m

Magnitude of charges = q= ?

Torque on electric dipole is given by formula

$\tau = pE \,sin\,theta\\\\p=q.r\\\\\theta = 90^o\\\\\tau=qrE\,sin\,(90)\\\\\tau=qrE\\\\q=\frac{\tau}{rE}\\\\q=\frac{6.6\times 10^{-26}}{(1.1\times 10^{-10}) (8.5\times 10^{4})}\\\\q=0.705\times 10^{-20}\, C$

Answer 2

Answer: q = 7.06 × 10^-21 C

the magnitude of the charges that make up the dipole is 7.06 × 10^-21 C

Explanation:

Given:

Torque, τ = 6.60×10^−26N⋅m

Angle made by p with a uniform electric field, θ = 90° (perpendicular)

Electric field, E = 8.50×10^4N/C

Length between dipole r = 1.10×10^−10 m

Torque acting on the dipole is given by the relation,

τ = pE sinθ....1

But,

p = qr .....2

Substituting equation 1 to 2

τ= qrEsinθ ....3

Making q the subject of formula

q = τ/rEsinθ .....4

Where;

q = magnitude of the charges that make up the dipole.

Substituting the given values into equation 4:

q = 6.60×10^−26N⋅m/(1.10×10^−10 m × 8.50×10^4N/C × sin90°)

q = 0.70588 × 10^-20 C

q = 7.06 × 10^-21 C