An electric dipole placed in an electric field of intensity 2 × 105 N/C at an angle of30° experiences a torque equal to 4 Nm. The charge on the dipole of dipole length2 cm is(a) 7 µC (b) 8 mC (c) 2 mC (d) 5 mC
Question
An electric dipole placed in an electric field of intensity 2 × 105 N/C at an angle of30° experiences a torque equal to 4 Nm. The charge on the dipole of dipole length2 cm is(a) 7 µC (b) 8 mC (c) 2 mC (d) 5 mC
Solution
The torque τ experienced by an electric dipole in an electric field is given by the formula:
τ = pEsinθ
where: τ is the torque, p is the dipole moment, E is the electric field intensity, and θ is the angle between the dipole moment and the electric field.
We can rearrange this formula to solve for the dipole moment p:
p = τ / (Esinθ)
Substituting the given values:
p = 4 Nm / (2 × 10^5 N/C * sin(30°)) p = 4 Nm / (1 × 10^5 N/C) p = 4 × 10^-2 Cm
The dipole moment p is also equal to the product of the charge q on the dipole and the dipole length d:
p = qd
We can rearrange this formula to solve for the charge q:
q = p / d
Substituting the given values:
q = 4 × 10^-2 Cm / 2 cm q = 2 × 10^-2 C
Therefore, the charge on the dipole is 2 × 10^-2 C, or 20 mC. This is not one of the options given, so there may be a mistake in the problem or the options.
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