How much charge does a 9.0 V battery transfer from the negative to the positive terminal while doing 27 J of work?

Verified step by step guidance

1. Understand the relationship between work, charge, and voltage. The formula to use here is: $W=qV$, where $W$ is the work done (in joules), $q$ is the charge (in coulombs), and $V$ is the voltage (in volts).
2. Rearrange the formula to solve for charge $q$. This gives: $q=\frac{W}{V}$.
3. Substitute the given values into the formula. Here, $W=\; 27$ J and $V=\; 9.0$ V.
4. Perform the division to calculate the charge $q$. This step involves dividing the work by the voltage: $q=\frac{27}{9.0}$.
5. The result will give you the charge transferred in coulombs. Ensure the units are consistent and verify the calculation for accuracy.

Key Concepts

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is measured in coulombs (C) and can be positive or negative. In the context of a battery, the movement of charge from the negative terminal to the positive terminal is what generates electric current.

Voltage

Voltage, or electric potential difference, is the measure of the energy per unit charge available to move charges between two points in an electric field. It is measured in volts (V). In this question, the 9.0 V battery indicates the potential difference that drives the charge movement, which is essential for calculating the amount of charge transferred.

Work-Energy Principle

The work-energy principle states that the work done on an electric charge by an electric field is equal to the change in electric potential energy of the charge. In this scenario, the battery does 27 J of work, which can be related to the charge transferred using the formula W = Q × V, where W is work, Q is charge, and V is voltage.