Question

An electron is carried from the positive terminal to the negative terminal of a 9 v battery. How much work is required in carrying this electron?

Answer 1

The work required to carry an electron from the positive terminal to the negative terminal of a 9 V battery is 4.4 × 10⁻¹⁸ joules.

Step-by-step explanation:

The work required to carry an electron from the positive terminal to the negative terminal of a 9 V battery can be calculated using the relationship between work (W), charge (q), and voltage (V). The work done on a charge is equal to the product of the charge and the potential difference (voltage) it moves through. The charge of an electron is – 1.6 × 10⁻¹⁹ C. The voltage of the battery is 9 V. Applying the formula W = qV, we get:

W = (1.6 × 10⁻¹⁹ C) × (9 V) = –4.4 × 10⁻¹⁸ J

The negative sign indicates that work is done against the electric field. However, when referring to the magnitude of work done, we can say that it is 4.4 × 10⁻¹⁸ joules.

Answer 2

The work required for the electron is
`1.44x10^(-18) J`

Step-by-step explanation:

The work of the electron is calculated in the equation:

`W= e * V`

`e= 1.6x10^(-19) C`

C= Coulomb = s*A

V= Volts =
`(W)/(A)`

W=
`(J)/(s)`

`W= 1.6 x10^(-19)C *9 V\\W=1.44x10^(-18) C*V\\ W=1.44x10^(-18) s*A*(W)/(A) \\W=1.44x10^(-18) W*s\\W=1.44x10^(-18) (J)/(s)*s \\W=1.44x10^(-18) J\\`