Question

Two identical pendulums A and B are suspended from the one point. The Bob's A and B are given positive charges Q1 and Q2 respectively. If mass of A and B are m1 and m2 (m1>m2) and angles from vertical of A and B are theta1 and theta2 respectively​

Answer 1

Answer:

angle made with the vertical by pendulum A is smaller than angle made with the vertical by pendulum B

Explanation:

Let the two pendulum is suspended from same point and at equilibrium the separation between them is "d"

so the horizontal force between them is given as

$F_x = \frac{kQ_1Q_2}{d^2}$

above is the horizontal force on both of the pendulums

Now the vertical force on both of the pendulums

$F_y = mg$

now the angle made by the first string with vertical is given as

$tan\theta_1 = \frac{F_x}{F_y}$

$tan\theta_1 = \frac{kQ_1Q_2}{d^2 m_1g}$

angle made by 2nd string with the vertical

$tan\theta_2 = \frac{kQ_1Q_2}{d^2 m_2g}$

as we know that

$m_1 > m_2$

so we will have

$\theta_1 < \theta_2$

#Learn

Topic : Electrostatic force

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