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Question: An electric Dipole of length \[10cm\] having charges \[ \pm 6 \times {10^{ - 3}}C\] placed at \[{30^...

An electric Dipole of length 10cm10cm having charges ±6×103C \pm 6 \times {10^{ - 3}}C placed at 30o{30^o} w.r.t a uniform electric field experiences a torque of magnitude 63Nm6\sqrt 3 Nm. Calculate
(1) Magnitude of electric Field
(2) Potential energy of dipole.

Explanation

Solution

At first, we need to calculate the Dipole moment of the dipole. From the obtained value we can get the value of potential energy. Using the equation of torque, the value of the electric field can be obtained.

Formula Used:
1. p=qdp = q \cdot d
Where,
p=\vec p = Dipole moment
d=d = Distance between the dipoles
q=q = Charge of the dipole
2. τ=p×E=pEsinθ\tau = \vec p \times \vec E = pE\sin \theta
Where,
τ=\tau = Torque
Dipole Moment
E=\vec E = Electrical Field
3. U=pE=pEcosθU = \vec p \cdot \vec E = pE\cos \theta
p=\vec p = Dipole Moment
E=\vec E = Electric Field
U=U = Potential Energy

Complete step-By-Step Answer:
Given:
Distance between the dipoles =10cm = 10cm denoted by dd
Charge of the dipole=±6×103C = \pm 6 \times {10^{ - 3}}C denoted by qq
Therefore, dipole moment is given by:
p=qdp = q \cdot d
Where,
p=\vec p = Dipole moment
d=d = Distance between the dipoles
q=q = Charge of the dipole
p=±6×103C×102p = \pm 6 \times {10^{ - 3}}C \times {10^{ - 2}}
Make sure, centimeter is converted to meters. So substituting the values of charge and distance we get,
p=6×104Cmp = 6 \times {10^{ - 4}}Cm
1. Now, calculating the value of electrical field from the equation of torque, by putting value of dipole moment in the torque equation:
τ=p×E=pEsinθ\tau = \vec p \times \vec E = pE\sin \theta
Where,
τ=\tau = Torque
p=\vec p = Dipole Moment
θ=30o\theta = {30^o}
Given, τ=63Nm\tau = 6\sqrt 3 Nm
Substituting the known values in the equation:
63Nm=6×104Cm×E6\sqrt 3 Nm = 6 \times {10^{ - 4}}Cm \times \vec E
E=23N/C\vec E = 2\sqrt 3 N/C
2. Now, putting the value of electrical field in the equation of potential energy:
U=pE=pEcosθU = \vec p \cdot \vec E = - pE\cos \theta
p=\vec p = Dipole Moment
E=\vec E = Electric Field
U=U = Potential Energy
U=(6×104)×(23×104)×cos60oU = - (6 \times {10^{ - 4}}) \times (2\sqrt 3 \times {10^4}) \times \cos {60^o}
U=18J\Rightarrow U = - 18J
Potential Energy is denoted by a negative sign, the charge is moved against the direction of the electric field.
This is our required solution.

Note: Electrical Dipole Moment is a quantity of measure of separation of two opposite charges separated by a distance. Dipole moment is a vector quantity.
Electric field can be defined as force exerted per unit charge. This is also a vector quantity.
Potential Energy can be defined as the amount of energy required to move a charge against the electric field.