Question

An electric dipole is placed at angle of 30 degree with an electric field of 2×10^5Nc^-1. It's length is 2Cm and if it experience a torque about 4Nm then it's charge is _______Mc.​

Answer 1

Answer:. 10Mc

Explanation:

WE KNOW THAT

Torque=PE sin theta-----(1)

& We also know that P=q.2L

So, By putting P in-----(1)

We get

Torque=q.2LE sin theta

Now,

q=torque/2LE sinTheta

So, by putting values

q=4/(2×2×10^-2×2×10^-5 sin30°)

So,,,,q=10^7C=10MC

Hope it helps u

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Answer 2

• Electric field intensity, E = 2 × 10⁵ NC⁻¹
• Length, l = 2 cm = 0.02 m
• Torque experience, τ = 4 Nm
• Angle, θ = 30°

We know,

$\sf{\overrightarrow{t}=\overrightarrow{P} \times \overrightarrow{E}}\\\\\sf{\implies \overrightarrow{t}=PE\ sin \theta}$

◙ For θ = 0°, τ = 0

◙ For θ = 30°,

$\sf{4=P \times 2 \times 10^5 \times sin 30^\circ}\\\\\sf{\implies 4=P \times 10^5 \times \dfrac{1}{2} \times 2 }\\\\\sf{\implies 4=P \times 10^5}\\\\\sf{\implies P=\dfrac{4}{10^5} }\\\\\sf{\implies P=4 \times 10^-^5}$

We also know that,

$\sf{Charge=\dfrac{P}{l} }$

Putting the values :-

$\sf{\implies q=\dfrac{4 \times 10^-^5}{0.02} }\\\\\sf{\implies q=\dfrac{4 \times 10^-^5}{2 \times 10^-^2} }\\\\\sf{\implies q=2 \times 10^{-5+2}}\\\\\sf{\implies q=2\times10^{-3}\ C}\\\\\boxed{\sf{\implies q=2\ mC}}$