Question

An electic dipole of length 2cm is placed with its axis making an angle 60 degree with respect to a uniform electric field of 105 N/C. If it experiences a torque of 13.856 Nm, calculate the- magnitude of the charge on the dipole & PE of the dipole.

Answer 1

we know that,
Torque acting on a dipole = pEsin(angle)
$13.856 =q \times 0.02\times 105 \times \sin(60) \\q = \frac{1 3.856 \times 2}{105 \times \sqrt{3} \times 0.02 } = 7.618c$

Answer 2

Answer: The magnitude of the charge is 8 m-C and the potential energy is 16 J.

Explanation:

Given that,

Length l = 2 cm

Electric field $E=105 N/C$

Torque $\tau=13.856 Nm$

Angle $\theta=60^{0}$

We know that,

$\tau = qlEsin\ \theta$

$q = \dfrac{\tau}{lE\ sin\ \theta}$

$q=\dfrac{13.856\ N-m\times 2}{2\times10^{-2}\ m\times10^{5} N/C \sqrt{3}}$

$q = 0.008\ C$

$q = 8\ mC$

The potential energy of the dipole

$U = pE\ cos\ \theta$

$U = q2lE cos\ \theta$

$U= 8\times10^{-3}\times 2\times0.02\ m\times10^{5}\ N/C\times cos60^{0}$

$U= 8\times10^{-3}\times 2\times0.02\ m\times10^{5}\ N/C\times\dfrac{1}{2}$

$U=16 J$

Hence, The magnitude of the charge is 8 m-C and the potential energy is 16 J.