Question:

If the energy of a 100 $\mu$F capacitor charged to 6 kV could all be used to lift a 50 kg mass, what would be the greatest vertical height through which mass could be raised?

Updated On: Jun 23, 2023

0.6 mm
3.6 m
1.2 mm
12 m

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The Correct Option is B

Solution and Explanation

Energy stored in the capacitor is used in lifting the mass. $\therefore \hspace25mm mgh = \frac{1}{2}CV^2$ $ \, \, 50 \times 9.8 \times h = \frac{1}{2} \times 100 \times 10^{-6} \times (6 \times 10^2)^2$ $\, \, 50 \times 9.8 \times h = \frac{1}{2} \times 100 \times 10^{-6} \times 36 \times 10^6$ $\therefore \hspace20mm h = \frac{1800}{50 \times 9.8} = 3.6 \, m$

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Notes on electrostatic potential and capacitance

Concepts Used:

Electrostatic Potential and Capacitance

Electrostatic Potential

The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.

Some major things that we should know about electric potential:

They are denoted by V and are a scalar quantity.
It is measured in volts.

Capacitance

The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.

The capacitor is in Series and in Parallel as defined below;

In Series

Both the Capacitors C₁ and C₂ can easily get connected in series. When the capacitors are connected in series then the total capacitance that is C_totalis less than any one of the capacitor’s capacitance.

In Parallel

Both Capacitor C₁ and C₂ are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is C_total is any one of the capacitor’s capacitance.