An electric dipole of magnitude 17 × 10-8 C - m is aligned in a uniform electric field of 2 × 105 N / C. What is value of potential energy of the dipole to hold the dipole at 30° with the direction of electric field?

Question

An electric dipole of magnitude 17 × 10-8 C - m is aligned in a uniform electric field of 2 × 105 N / C. What is value of potential energy of the dipole to hold the dipole at 30° with the direction of electric field? (A) -1.7 × 10-3 J (B) -1.7 × 10-2 J (C) -2.9 × 10-2 J (D) 5.1 × 10-2 J

Tardigrade Admin · Accepted Answer

P E =-P E cos θ =-17 × 10-8 × 2 × 105 × cos 30° =-17 √3 × 10-3 =-2.9 × 10-2 J

Q. An electric dipole of magnitude $17 \times 1 0^{- 8} C - m$ is aligned in a uniform electric field of $2 \times 1 0^{5} N / C$ . What is value of potential energy of the dipole to hold the dipole at $3 0^{\circ}$ with the direction of electric field?

$- 1.7 \times 1 0^{- 3} J$

$- 1.7 \times 1 0^{- 2} J$

$- 2.9 \times 1 0^{- 2} J$

$5.1 \times 1 0^{- 2} J$

$PE = - PE cos θ$
$= - 17 \times 1 0^{- 8} \times 2 \times 1 0^{5} \times cos 3 0^{\circ}$
$= - 173 \times 1 0^{- 3}$
$= - 2.9 \times 1 0^{- 2} J$

Q. An electric dipole of magnitude 17×10−8C−m is aligned in a uniform electric field of 2×105N/C. What is value of potential energy of the dipole to hold the dipole at 30∘ with the direction of electric field?

−1.7×10−3J

−1.7×10−2J

−2.9×10−2J

5.1×10−2J