SATHEE: Electrostatics Question 246

Electrostatics Question 246

Question: A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at $x = + 1$ cm and C be the point on the y-axis at $y = + 1$ cm. Then the potentials at the points A, B and C satisfy [IIT-JEE (Screening) 2001]

Options:

A) ${(\frac{q^{2} L}{2 π ε_{0} m g})}^{\frac{1}{3}}$

B) ${(\frac{q L^{2}}{2 π ε_{0} m g})}^{\frac{1}{3}}$

C) ${(\frac{q^{2} L^{2}}{4 π ε_{0} m g})}^{\frac{1}{3}}$

D) ${(\frac{q^{2} L}{4 π ε_{0} m g})}^{\frac{1}{3}}$

Show Answer

Answer:

Correct Answer: A

Solution:

In equilibrium $F_{e} = T sin θ$ ……. (i) $m g = T cos θ$ ……. (ii) $\tan θ = \frac{F_{e}}{m g} = \frac{q^{2}}{4 π ε_{o} x^{2} \times m g}$ also $\tan θ \approx \sin θ = \frac{x / 2}{L}$ Hence $\frac{x}{2 L} = \frac{q^{2}}{4 π ε_{o} x^{2} \times m g}$

therefore $x^{3} = \frac{2 q^{2} L}{4 π ε_{o} m g}$

therefore $x = {(\frac{q^{2} L}{2 π ε_{o} m g})}^{1 / 3}$

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Electrostatics Question 246

Question: A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x=+1 cm and C be the point on the y-axis at y=+1 cm. Then the potentials at the points A, B and C satisfy [IIT-JEE (Screening) 2001]

Options:

Answer:

Solution:

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Question: A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at $x = + 1$ cm and C be the point on the y-axis at $y = + 1$ cm. Then the potentials at the points A, B and C satisfy [IIT-JEE (Screening) 2001]