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How much work is required to carry an electron from positive terminal of 12Volt battery to negative terminal?

User Southoz
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Final answer:

The work required to carry an electron from the positive terminal to the negative terminal of a 12V battery is 1.922 x 10^-18 joules. This is calculated using the formula W = qV, with the charge of an electron being -1.602 x 10^-19 coulombs.

Step-by-step explanation:

To determine how much work is required to carry an electron from the positive terminal to the negative terminal of a 12V battery, we use the relationship between work (W), electric potential difference (V), and charge (q).

The formula for work done in moving a charge in an electric field is W = qV, where W is the work in joules, q is the charge in coulombs, and V is the electric potential difference in volts.

An electron has a charge of approximately -1.602 x 10-19 coulombs (e). Substituting the given values, we have W = (-1.602 x 10-19 C)(12 V), which yields W = -1.922 x 10-18 joules. The negative sign indicates that the work is done by the electric field on the electron, but in terms of magnitude, the work required is 1.922 x 10-18 joules.

User Jeremy Andrews
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