Two charges of magnitude -4C and 4C are placed at points (1, 2, 3) and (3, 4, 2) respectively. What is the dipole moment?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
The dipole moment will be in the direction of the vector connecting -4C to 4C i.e., from (1, 2, 3) to (3, 4, 2).
Let $\to L \to$ which is a vector representing the distance between -4C charge to +4C Charge
$\to L$ = Portion vector of +4C charge – Position vector of -4C charge.

$\to L$ = $(3 ˆ i + 4 ˆ j + 2 ˆ k) - (1 ˆ i + 2 ˆ j + 3 ˆ k)$
$= 2 ˆ i + 2 ˆ j - ˆ k$
Now $\to p$ = $q \to L = 4 \times (2 ˆ i + 2 ˆ j - k)$
$= 8 ˆ i + 8 ˆ j - 4 ˆ k$

Suggest Corrections

Similar questions

_{1} = ˆ i + 2 ˆ j + ˆ k

a n d

_{2} = 2 ˆ i - 3 ˆ j + ˆ k

If \vec{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k} and \vec{c} = \hat{i} + 2 \hat{j} - \hat{k}, find |3 \vec{a} - 2 \vec{b} + 4 \vec{c}| .

+ q, - 2 q

+ q

(0, 1, 0), (0, 0, 0)

(1, 0, 0)

Join BYJU'S Learning Program