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 E⃗ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ____

Simple pendulum with charged bob in horizontal electric field tension calculation

Q. 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 _____.

A. $mg - qE$

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

C. $mg + qE$

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

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

Explanation

When the electric field is applied horizontally, two forces act on the bob:

• Gravitational force acting vertically downward = $mg$
• Electric force acting horizontally = $qE$

At equilibrium, the bob comes to rest making an angle with the vertical such that the tension in the string balances both forces.

The tension T acts along the string and is the resultant of the gravitational and electric forces.

Since $mg$ and $qE$ act perpendicular to each other, the magnitude of the resultant force is:

$$ T = \sqrt{(mg)^2 + (qE)^2} $$

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

Therefore, the final tension in the string is $\sqrt{m^2g^2 + q^2E^2}$.

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