The electric potential inside a living cell is lower than the potential outside. Suppose the electric potential difference between the inner and the outer cell wall is 0.095 V, a typical value. To maintain the internal electrical balance, the cell pumps out sodium ions. How much work must be done to remove a single sodium ion (charge e)

The electric potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.The electric potential difference will be 1.52×10⁻²⁰J.

What is the electric potential difference ?

The electrical potential difference is the amount of effort done in an electrical field to shift a unit charge from one spot to another.

Traditional current flows from positive to negative terminals, signifying positive charge transfer in that direction.

The given data in the problem is

is the electric potential difference between the inner and the outer cell wall = 0.095 V

is the amount of work done

q is the charge on the electron =1.6×10⁻¹⁹C

Electric potential difference is given by the formula

Hence the electric potential difference will be 1.52×10⁻²⁰J.

To learn more about the electric potential difference refer to the link;

link

1.52 × 10⁻²⁰ J

Explanation:

The electrical potential difference is defined as the amount of work done in carrying a unit charge from one point to another point in an electric field. Electric potential difference is measured in volts. It is given by the formula:

ΔV = ΔU / q

ΔV is electric potential difference between the two points, ΔU is the work done and q is the unit charge.

Given that ΔV = 0.095 V, q = 1.6 × 10⁻¹⁹ C. Hence:

ΔU = ΔV.q

ΔU = 0.095 V * 1.6 × 10⁻¹⁹ C

ΔU = 1.52 × 10⁻²⁰ J

where are the vectors ?