Question

As a particle moves 10.0 m along an electric field of strength 75 n/c, its electrical potential energy decreases by 4.8 × 10−16 j. what is the particle's charge? 2. what is the potential difference between the initial and final locations of the particle in problem 1?

Answer 1

Part 1)

let say the force on the charge particle which is placed in electric field is given as

$F = qE$

now work done to move a charge from one position to other position is known as change in electrostatic potential energy

So here we will have

$\Delta U = W$

$\Delta U = F.d$

$\Delta U = qE. d$

$4.8 \times 10^{-16} = q(75 N/C) (10)$

$q = 6.4 \times 10^{-19} C$

Part 2)

change in potential energy of charge is product of charge and potential difference between two points

$\Delta U = q\Delta V$

so here we will have

$4.8\times 10^{-16} J = 6.4\times 10^{-19} (\Delta V)$

$\Delta V = 750 Volts$

Answer 2

a. The charge of the particle is $6.4 \times 10^{-19}$ Coulombs.

b. The potential difference between the initial and final locations of the particle in problem 1 is 750 Volts.

Given the following data:

• Distance = 10 meters

• Electric field strength = 75 N/C

• Change in electrical potential energy = $4.8 \times 10^{-16}$ Joules

To find the charge of the particle:

The change in electrical potential energy is the work done in moving a charge from one point to another.

Mathematically, the change in electrical potential energy is given by the formula:

$\Delta U =W = Fd$

Where:

• $\Delta U$ is the change in electrical potential energy.

• F is the force.

• d is the distance.

But, $F = qE$

$\Delta U = qEd$

Substituting the given parameters into the above formula, we have;

$4.8 \times 10^{-16} = q(75 \times 10)\\\\4.8 \times 10^{-16} = 750q\\\\q = \frac{4.8 \times 10^{-16}}{750}$

Charge, q = $6.4 \times 10^{-19}$ Coulombs.

2. To find the potential difference between the initial and final locations of the above particle:

$\Delta V = \frac{\Delta U}{q} \\\\\Delta V = \frac{4.8 \times 10^{-16}}{6.4 \times 10^{-19}}$

Potential difference = 750 Volts

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