Question

a dipole is present in an electrostatic field of magnitude 10^6 C if the work done in rotating it from its position of stable equilibrium to its position of unstable equilibrium equals f 2 * 10^(-23)J then find the magnitude of the dipole moment of this dipole ?​

Answer 1

Explanation:

Given

Work done (W) = $\bold{\mathsf{2 \times 10^{-23} J }}$

E (electric field) = $\bold{\mathsf{10^{6} N/C}}$

As we know that

Magnitude of work done (W) = 2pE

Put in the given data

$\bold{\mathsf{2 \times 10^{-23}=2p \times 10^{6}}}$

$\implies{\mathsf{p = \frac{2 \times 10^{-23}}{2 \times 10^{6}}}}$

By solving ,

$\implies{\mathsf{P = 10^{-29}}}$

Thus, the magnitude of the dipole movement of the dipole is $\implies{\mathsf{10^{-29}}}$C-m