Question

*＊✿❀Subh Sandhya mitron.....❀✿＊*



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Answer 1

${\large{\sf{\pink{\underline{\underline{\bigstar{ \: \: Given : }}}}}}}$

• Rotating and electric dipole of dipole movement 3.2 × 10^(-8) Cm.
• Uniform electric field of intensity 10⁴ N/C.

${\large{\sf{\pink{\underline{\underline{\bigstar{ \: \: To \: find : }}}}}}}$

• Amount of work done in the rotating and electric dipole.

${\large{\sf{\pink{\underline{\underline{\bigstar{ \: \:Solution : }}}}}}}$

Here we have :-

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bull \: \: { \sf{P=3×10 ^{−8}Cm}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bull \: \: { \sf{E=10^{4} \: \: N/C}}$

${ \sf{At \: stable \: equilibrium \: ( θ_1 )=0°}}$

${ \sf{At \: unstable \: equilibrium \: (θ_2 )=180°}}$

So,

${ \sf{ \large{Formula : - }}}\\ { \boxed{\large{ \green{ \sf{W=PE(cosθ _1 −cosθ _2 )}}}}}$

Work done in a rotating dipole is :-

${ \implies{\sf{(3 \times 10 ^{−8} )(10⁴ )[cos0° −cos180° ]}}}$

${\sf{\implies{3×10 ^{−4} [1−(−1)]}}}$

${\sf{\implies{3×10 ^{−4} \times (2)}}}$

${ \red{ \boxed{ \large{\sf{\implies{6×10 ^{−4} \: J}}}}}}$