Question

4. An electric dipole is placed at an
alignment angle of 30° with an electric
field of 2 X 105 N C. It experiences a
torque equal to 8 Nm. The charge on
the dipole if the dipole length is 1 cm is​

Answer 1

HEYA GM ❤

★» TORQUE (T)= Dipole(p) × field × Sin∅

DIPOLE (P) = q(charge) × 2L(length)

SO ;

» T = pEsin∅

» T = pEsin∅» 8 = q × 2L × 2 × 10^5 × Sin30°

» T = pEsin∅» 8 = q × 2L × 2 × 10^5 × Sin30° » 8 = q × 2(1) × 2 × 10^5 × 1/2

» T = pEsin∅» 8 = q × 2L × 2 × 10^5 × Sin30° » 8 = q × 2(1) × 2 × 10^5 × 1/2 » 8 = q × 2 × 10^5

» T = pEsin∅» 8 = q × 2L × 2 × 10^5 × Sin30° » 8 = q × 2(1) × 2 × 10^5 × 1/2 » 8 = q × 2 × 10^5 » q = 8 / 2 × 10^5

» T = pEsin∅» 8 = q × 2L × 2 × 10^5 × Sin30° » 8 = q × 2(1) × 2 × 10^5 × 1/2 » 8 = q × 2 × 10^5 » q = 8 / 2 × 10^5 » q = 4 × 10^-5 Cuolumb

A. HENCE ; CHARGE ON DIPOLE IS 4 × 10^-5 C.

TQ ❤ ❤

Answer 2

Answer:

The charge on the dipole is equal to 8mC, if length of the dipole is equal to 1cm.

Explanation:

We have given,

The length of the dipole, L = 1cm = 0.01m

Consider that the charge on the dipole is equal to 'q'.

Dipole moment of the dipole is given by:

Dipole moment = Dipole length × charge

$P = L*q$

$P=0.01q$

Torque acting on the dipole is given by:

$\vec \tau=\vec P$×$\vec E$

where E is electric field.

$\tau =PEsin\theta$                                                                          .............(1)

The angle of alignment of dipole, θ = 30°, E = 2×10⁵N/C and τ = 8Nm in equation (1);

$8=0.01*q*2*10^{5}\;sin30^o$

$q=8*10^{-3}C$

$q=8mC$

Therefore, the charge on the electric dipole is equal to 8 microcoulomb.