Question

The potential difference between two points, A and B, in an electric field is 2.00 volts. The energy required to move a charge of 8x10^-19 coulomb from point A to point B is

Answer 1

Answer:

$W_A_B=-1.6\times 10^{-18} J$

Explanation:

Let A and B be two points located in a uniform electric field, A being a distance d from B in the direction of the field. The work that an external force must do to bring a unit positive charge q from the reference point to the point considered against the electric force at constant speed, mathematically is expressed by:

$V_B_A=\frac{W_A_B}{q}$

Therefore, isolating $W_A_B$ and replacing the data provided:

$W_A_B=V_B_A *q=-2*(8\times 10^{-19}) =-1.6\times 10^{-18}J$

Answer 2

The energy required to move the given charge from point A to point B is of  $1.6 \times 10^{-18} \;\rm J$.

Given data:

The potential difference between the points A and B is, V' = 2.00 V.

The magnitude of charge is, $q = 8 \times 10^{-19}\;\rm C$.

The given problem is based on the work done to shift a charge from one point to another.

• When a charge is moved between the two points, then some work is to be done, which gets stored in the form of electrostatic potential energy.
• And the mathematical expression for the electrostatic potential energy is,

        W = U

        W = q × V'

       Solve by substituting values as,

       $W = (8 \times 10^{-19}) \times 2.00\\\\W = 1.6 \times 10^{-18} \;\rm J$

Thus, we can conclude that the energy required to move the given charge from point A to point B is of  $1.6 \times 10^{-18} \;\rm J$.

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