Question 291: Electrostats & Capacitance

Question

A simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is $\_\_\_\_$ .

Accepted Answer

When the bob is at its new equilibrium position, it is stationary. This means the net force acting on it is zero.
The free body diagram of bob is,
The gravitational force acting vertically downwards. Its magnitude is $\mathrm{F}_{\mathrm{g}}=\mathrm{mg}$.
Because the bob has a charge $q$ and is in a horizontal electric field $\vec{E}$, it experiences a horizontal force. Its magnitude is $\mathrm{F}_{\mathrm{e}}=\mathrm{qE}$. Let's assume it points to the left.
The force exerted by the string on the bob, acting along the length of the string towards the pivot point with magnitude T.
Let the string make an angle $	heta$ with the vertical when it reaches this new equilibrium position.
On resolving the forces into vertical and horizontal components.
Along vertical direction ( y - axis):
The upward component of the tension must balance the downward weight of the bob.
Therefore: $\mathrm{T} \cos 	heta=\mathrm{mg} \ldots$ (i)
Along horizontal direction ( $\mathrm{x}-$ axis): The horizontal component of the tension must balance the outward pull of the electric force.
Therefore : $\mathrm{T} \sin 	heta=\mathrm{qE} \ldots$ (ii)
Squaring and adding both the equations,
$$ \mathrm{T}^2 \sin ^2 	heta+\mathrm{T}^2 \cos ^2 	heta=\mathrm{q}^2 \mathrm{E}^2+\mathrm{m}^2 \mathrm{~g}^2 $$
$$\Rightarrow $$ $\mathrm{T}^2\left(\sin ^2 	heta+\cos ^2 	hetaight)=\mathrm{q}^2 \mathrm{E}^2+\mathrm{m}^2 \mathrm{~g}^2$
$$\Rightarrow $$ $\mathrm{T}^2=\mathrm{q}^2 \mathrm{E}^2+\mathrm{m}^2 \mathrm{~g}^2$
$$\Rightarrow $$ $$ \mathrm{T}=\sqrt{\mathrm{m}^2 \mathrm{~g}^2+\mathrm{q}^2 \mathrm{E}^2} $$
Therefore, the magnitude of the tension is $T=\sqrt{m^2 g^2+q^2 E^2}$.
Hence, the correct option is (B).

Answer

$m g-q E$

Answer

$\sqrt{m^2 g^2+q^2 E^2}$

Answer

$m \mathrm{~g}+q E$

Answer

$\sqrt{m^2 g^2-q^2 E^2}$

Electrostats & Capacitance

A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field \vec{E} is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is \_\_\_\_ .

Choose the correct answer: