What is the magnitude of electric field intensity due to a dipole of moment 4 × 10-8 C - m at a point distant 0.5 m from the center of dipole when line joining the point to the center of dipole makes an angle of 60° with dipole axis?

Question

What is the magnitude of electric field intensity due to a dipole of moment 4 × 10-8 C - m at a point distant 0.5 m from the center of dipole when line joining the point to the center of dipole makes an angle of 60° with dipole axis? (A) 4 × 10-5 N / C (B) 5 × 10-11 N / C (C) 2 × 107 N / C (D) 3.8 × 103 N / C

Tardigrade Admin · Accepted Answer

E =( P /4 π ε 0 r 3) √3 cos 2 θ+1 E =(4 × 10-8 × 9 × 109/0.5 × 0.5 × 0.5) × √3(0.5)2+1 E =(360/0.125) × √1.75 E =(360 × 1.32/0.125)=3801.6 =3.8 × 103 N / C

Q. What is the magnitude of electric field intensity due to a dipole of moment $4 \times 1 0^{- 8} C - m$ at a point distant $0.5 m$ from the center of dipole when line joining the point to the center of dipole makes an angle of $6 0^{\circ}$ with dipole axis?

$4 \times 1 0^{- 5} N / C$

$5 \times 1 0^{- 11} N / C$

$2 \times 1 0^{7} N / C$

$3.8 \times 1 0^{3} N / C$

$E = \frac{P}{4 π ε _{0} r ^{3}} 3 cos^{2} θ + 1$
$E = \frac{4 \times 1 0 ^{- 8} \times 9 \times 1 0 ^{9}}{0.5 \times 0.5 \times 0.5} \times 3 (0.5)^{2} + 1$
$E = \frac{360}{0.125} \times 1.75$
$E = \frac{360 \times 1.32}{0.125} = 3801.6$
$= 3.8 \times 1 0^{3} N / C$

Q. What is the magnitude of electric field intensity due to a dipole of moment 4×10−8C−m at a point distant 0.5m from the center of dipole when line joining the point to the center of dipole makes an angle of 60∘ with dipole axis?

4×10−5N/C

5×10−11N/C

2×107N/C

3.8×103N/C