1. Tardigrade
  2. Question
  3. Physics
  4. Consider a dipole with dipole moment 4.8 × 10-30 cm. Let a point P lies 4.1 × 10-9 m away from centre of dipole. Match the values of potentials given in column-II with column-I and mark the correct choice from the given codes. Column I Column II i Point is along axis of dipole nearer to positive charge. p -1.8 × 10-3 V ii Point is 45° above the axis but nearer to positive charge. q 1.8 × 10-3 V iii Point is 45° above the axis but nearer to negative charge. r 2.6 × 10-3 V Codes:

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Q. Consider a dipole with dipole moment $4.8 \times 10^{-30} cm$. Let a point $P$ lies $4.1 \times 10^{-9} m$ away from centre of dipole. Match the values of potentials given in column-II with column-I and mark the correct choice from the given codes.
Column I Column II
i Point is along axis of dipole nearer to positive charge. p $-1.8 \times 10^{-3} V$
ii Point is $45^{\circ}$ above the axis but nearer to positive charge. q $1.8 \times 10^{-3} V$
iii Point is $45^{\circ}$ above the axis but nearer to negative charge. r $2.6 \times 10^{-3} V$

Codes:

Electrostatic Potential and Capacitance

Solution:

(i) $V=\frac{9 \times 10^{9} \times 4.8 \times 10^{-30} \times \cos 0^{\circ}}{\left(4.1 \times 10^{-9}\right)^{2}}$
$=2.6 \times 10^{-3} V$
(ii) $V=\frac{k p \cos \theta}{r^{2}}$
$=1.8 \times 10^{-3} V$
(iii) $V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p \cos \theta}{r^{2}}$
$=\frac{9 \times 10^{9} \times 4.8 \times 10^{-30} \times \cos 135^{\circ}}{\left(4.1 \times 10^{-9}\right)^{2}}$
$= -1.8 \times 10^{-3} V$