Skip to content
Question of 67

Q.An electric dipole of 3×10−83\times10^{-8} coulomb-metre dipole moment is inclined at an angle of 30030^{0} to a uniform electric field and experiences a torque of 1.2×1031.2\times10^{3} newton-metre. Calculate

(i) Magnitude of electric field
(ii) Potential energy of the dipole.
(OR)
Three capacitors C1C_1, C2C_2 and C3C_3 are connected in series. Find the expression for their equivalent capacitance.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2023Subjective· 2mImportance★★★★★
0% · 0/67 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Torque on a dipole gives the field; potential energy follows from U=−pEcos⁡θU=-pE\cos\theta. (OR: series capacitors add as reciprocals.)

Torque on an electric dipole: τ=pEsin⁡θ\tau = pE\sin\theta, where pp is the dipole moment, EE the field, θ\theta the angle between them.

Given p=3×10−8p = 3\times10^{-8} C·m, θ=30∘\theta = 30^\circ, τ=1.2×103\tau = 1.2\times10^{3} N·m.

  1. E=τpsin⁡θ=1.2×1033×10−8×sin⁡30∘=1.2×1033×10−8×0.5=1.2×1031.5×10−8=8×1010 N/CE = \dfrac{\tau}{p\sin\theta} = \dfrac{1.2\times10^{3}}{3\times10^{-8}\times\sin30^\circ} = \dfrac{1.2\times10^{3}}{3\times10^{-8}\times0.5} = \dfrac{1.2\times10^{3}}{1.5\times10^{-8}} = 8\times10^{10}\text{ N/C} Re-checking the arithmetic: 1.2×103/1.5×10−8=0.8×1011=8×10101.2\times10^{3}/1.5\times10^{-8} = 0.8\times10^{11} = 8\times10^{10} N/C.
  2. Potential energy of the dipole: U=−pEcos⁡θ=−(3×10−8)(8×1010)cos⁡30∘=−(2.4×103)(0.866)≈−2.08×103U = -pE\cos\theta = -(3\times10^{-8})(8\times10^{10})\cos30^\circ = -(2.4\times10^{3})(0.866) \approx -2.08\times10^{3} J. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.